The compound shown below is named cis-1,3-dichlorocyclohexane.
In the name "cis-1,3-dichlorocyclohexane," "cis" refers to the arrangement of the chlorine atoms on the cyclohexane ring. In the cis isomer, the two chlorine atoms are on the same side of the ring. The "1,3" indicates the positions of the chlorine atoms on the cyclohexane ring, with one chlorine attached to carbon atom 1 and the other chlorine attached to carbon atom 3.
The term "dichloro" indicates that there are two chlorine atoms in the molecule, and "cyclohexane" refers to the six-carbon ring structure.
Therefore, the proper name for the compound is cis-1,3-dichlorocyclohexane.
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What is the equivalent value in feet to \( 146 \mathrm{~cm} \) ? \( 1.03 \mathrm{ft} \) \( 0.432 \mathrm{ft} \) \( 62.2 \mathrm{ft} \) \( 4.79 \mathrm{ft} \)
The equivalent value in feet to 146 cm is 4.79 ft.
To convert centimeters (cm) to feet (ft), we need to use the conversion factor of 1 ft = 30.48 cm.
First, we divide 146 cm by 30.48 cm/ft to obtain the value in feet:
\( \frac{146 \, \text{cm}}{30.48 \, \text{cm/ft}} = 4.79 \, \text{ft} \)
Therefore, the equivalent value in feet to 146 cm is approximately 4.79 ft.
The conversion factor 1 ft = 30.48 cm is derived from the definition of the foot in the International System of Units (SI). It is equivalent to exactly 30.48 centimeters. By dividing the given value in centimeters by the conversion factor, we can convert it to the corresponding value in feet. In this case, 146 cm divided by 30.48 cm/ft gives us 4.79 ft, which is the equivalent length in feet.
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1.44 M aqueous solution of a weak base has a pH of 9.92 at 298
K. Calculate pKb for this weak base. Enter your
answer to 2 decimal places.
The pKb for the weak base in the 1.44 M aqueous solution, with a pH of 9.92 at 298 K, is approximately 4.08.
pKb is a measure of the strength of a base and is defined as the negative logarithm (base 10) of the equilibrium constant for the dissociation of the base. To calculate pKb, we first need to determine the concentration of the base.
From the given information, we know that the solution is 1.44 M, indicating the concentration of the weak base. Next, we use the pH value of 9.92 to determine the concentration of the hydroxide ion (OH⁻), which is formed from the dissociation of the weak base.
Using the equation for the dissociation of water, we can calculate the concentration of OH⁻. At 298 K, the concentration of OH⁻ is equal to 10(-pOH), where pOH is the negative logarithm (base 10) of the hydroxide ion concentration.
Since pH + pOH = 14, we have pOH = 14 - 9.92 = 4.08. Therefore, the concentration of OH⁻ is 10(-4.08).
Finally, we use the concentration of OH⁻ to calculate pKb. pKb = 14 - pOH = 14 - 4.08 ≈ 9.92.
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1. Briefly describe the hazards you should be aware of when you work with: (a) diethyl ether (b) 3MHCl 2. Briefly explain or describe the following: (a) How would you determine which layer is the aqueous layer after you add NaOH solution to the ether solution of your compounds? (b) What visible evidence(s) of reaction will you see when you acidify the NaOH extract with HCl solution? (c) In which layer would p-toluic acid be more soluble if p-toluic acid were added to a twolayer mixture of diethyl ether and water? (d) How would the results differ if you added sodium p-toluate instead of p-toluic acid to the two-layer mixture of diethyl ether and water? 3. How many milliliters of 3.0MHCI would be required to neutralize 30.mL of 0.50 NaOH ? (Show your work).
The volume of HCl required in milliliters is 4.2mL.
Hazards of Diethyl Ether Diethyl ether is a flammable and volatile organic compound that can pose a variety of risks to those who work with it. Because it is highly volatile, it has a high vapor pressure and can rapidly evaporate, causing dizziness and respiratory problems in those who inhale it. Diethyl ether can also cause skin irritation and dryness, as well as contact dermatitis. Because of its low boiling point, it is also possible that it could boil over or ignite if it is heated too quickly or if there is an open flame nearby.
Milliliters of 3.0M HCI Required To calculate the milliliters of 3.0MHCI required to neutralize 30.mL of 0.50 NaOH, the following equation is used: NaOH + HCl → NaCl + H2OFirst, the moles of NaOH are calculated:0.50 NaOH x (1 mole NaOH / 40 g/mol NaOH) = 0.0125 moles NaOH Next, the moles of HCl required are calculated using the mole ratio of NaOH to HCl:0.0125 moles NaOH x (1 mole HCl / 1 mole NaOH) = 0.0125 moles HCl Finally, the milliliters of 3.0MHCI required are calculated using the following equation: moles HCl = molarity HCl x volume HCl (in liters)0.0125 moles HCl = 3.0M x volume HCl (in liters)volume HCl (in liters) = 0.0125 moles HCl / 3.0M = 0.0042 liters Hence, the volume of HCl required in milliliters is 4.2mL.
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What is the pI of aspartic acid? Show the calculation and
explain what is meant by pI with pictures
The pI of aspartic acid is 2.77.
The pI, or isoelectric point, of a molecule is the pH at which the molecule carries no net electrical charge. To calculate the pI of aspartic acid, we need to consider its ionizable groups and determine the pH at which they are neutralized.
Aspartic acid has two ionizable groups: the carboxyl group (-COOH) and the amino group (-NH₂). At low pH values, the carboxyl group is protonated (COOH²⁺) and carries a positive charge, while the amino group is deprotonated (NH₂) and carries a negative charge.
At high pH values, the carboxyl group is deprotonated (COO⁻) and carries a negative charge, while the amino group is protonated (NH³⁺) and carries a positive charge.
The pI is the pH at which these ionizable groups are neutralized, resulting in a molecule with no net charge. In the case of aspartic acid, the pI can be calculated as the average of the pKa values of the two ionizable groups.
The pKa values for the carboxyl group of aspartic acid are approximately 2.0 and 4.0, while the pKa value for the amino group is around 9.9. Taking the average of the pKa values, we get:
pI = (pKa1 + pKa2) / 2 = (2.0 + 4.0) / 2 = 3.0
Therefore, the pI of aspartic acid is approximately 3.0.
Pictures can be helpful in understanding the concept of pI. Here's a visual representation:
1. At low pH (acidic conditions), the carboxyl group is protonated (COOH²⁺) and carries a positive charge, while the amino group is deprotonated (NH₂) and carries a negative charge. The molecule has an overall positive charge.
2. At high pH (basic conditions), the carboxyl group is deprotonated (COO⁻) and carries a negative charge, while the amino group is protonated (NH³⁺) and carries a positive charge. The molecule has an overall negative charge.
3. At the pI, the ionizable groups are neutralized, resulting in a molecule with no net charge. The carboxyl group and the amino group are in their deprotonated (or protonated) forms, and the molecule exists as a zwitterion.
The pI represents the pH at which the molecule is electrically neutral, and it is an important property for various biological and chemical applications.
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Please answer all parts of this question. Include relevent schemes,
structure, mechanism and explanation. Thank you.
From the molecules below (B, C and \( \mathbf{D} \) ), predict which molecule will give the most enantioselective reduction and the least enantioselective reduction. Which reduction will proceed most
Molecule D will give the most enantioselective reduction, while molecule C will give the least enantioselective reduction.
Enantioselective reduction refers to a reduction reaction that selectively produces one enantiomer over the other. Several factors can influence the enantioselectivity of a reduction reaction, including the presence of chiral catalysts or chiral environments.
1. Analysis of Molecules: Analyze the structures of molecules B, C, and D to determine their potential for enantioselective reduction. Look for chiral centers or asymmetry within the molecules.
2. Chiral Centers: Identify the presence of chiral centers in each molecule. Chiral centers are carbon atoms bonded to four different substituents, leading to two possible stereoisomers (enantiomers).
3. Enantioselective Catalysts: Consider the availability of chiral catalysts or chiral environments. Chiral catalysts can influence the stereochemistry of the reduction reaction and promote the formation of a specific enantiomer.
4. Molecular Structure: Assess the overall molecular structure and the presence of steric hindrance. Bulky substituents or groups in close proximity to the reactive site may affect the accessibility of the reducing agent, potentially leading to lower enantioselectivity.
Based on these considerations, molecule D is likely to give the most enantioselective reduction due to the presence of a chiral center and potentially favorable steric and electronic factors. On the other hand, molecule C may give the least enantioselective reduction, possibly due to the absence of a chiral center or unfavorable steric factors that hinder enantioselectivity.
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Show the reaction for the reaction of phenylmagnesium bromide with acetone, followed by acidic workup. Draw the structures NEATL. by hand. Be sure to use numbers to denote separate reaction steps. What experimental evidence do you have to support that the structure of the major organic product of the reaction is what you drew above? You need to cite specific data (TLC, IR, \& NMR).
The reaction of phenylmagnesium bromide with acetone followed by acidic workup produces 1-phenyl-2-propanol as the major organic product. Experimental evidence from techniques like TLC, IR, and NMR can confirm the structure of the product through comparison and analysis of its properties and spectra.
The reaction of phenylmagnesium bromide (PhMgBr) with acetone followed by acidic workup proceeds as follows:
Step 1: Addition of phenylmagnesium bromide (PhMgBr) to acetone
PhMgBr + acetone → PhMgC(O)CH₃
Step 2: Acidic workup
PhMgC(O)CH₃ + H₃O⁺ → PhCH(OH)CH₃ + MgBrOH
The major organic product of this reaction is PhCH(OH)CH₃, which is also known as 1-phenyl-2-propanol.
Experimental evidence to support the structure of the major organic product can be obtained from techniques such as TLC (Thin-Layer Chromatography), IR (Infrared Spectroscopy), and NMR (Nuclear Magnetic Resonance).
1. TLC: Thin-Layer Chromatography can be used to analyze the reaction mixture and compare the migration of the major product with reference compounds. The Rf value (retention factor) can help confirm the presence of the desired product.
2. IR: Infrared Spectroscopy can be used to identify functional groups present in the major organic product. The characteristic absorption peaks of hydroxyl (OH) and carbonyl (C=O) groups can be observed in the IR spectrum.
3. NMR: Nuclear Magnetic Resonance spectroscopy can provide detailed structural information about the major organic product. The NMR spectrum can confirm the presence of the phenyl group (Ph), the hydroxyl group (OH), and the protons of the methyl and methylene groups.
By comparing the experimental data obtained from TLC, IR, and NMR analysis of the major product with the expected spectra of 1-phenyl-2-propanol, it can be concluded whether the structure of the major organic product is consistent with the drawn structure.
It's important to note that the specific experimental evidence may vary depending on the actual analysis performed in a laboratory setting.
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Chloride [References] A 10.00 mL diluted chloride sample was titrated with 0.01246 M AgNO,, and 18.46 mL AgNO, was required to reach the endpoint. How would the following errors affect the calculated concentration of CI? a. The student read the molarity of AgNO, as 0.01426 M instead of 0.01246 M. The experimentally calculated moles of Ag would be too, so the calculated moles of CI would come out too [ the unknown would come out too b. The student took the initial buret reading correctly but took the final buret reading from the top of the miniscus. The experimentally determined moles of Ag would be too, so the calculated moles of CI would come out CI concentration. Submit Answer Try Another Version 2 item attempts remaining The calculated [CI] in as would the calculated
Errors in a chloride concentration experiment can affect the calculated [tex]Cl^-[/tex] concentration. Incorrectly reading the molarity of [tex]AgNO_3[/tex]overestimates the concentration, while taking the final buret reading from the top of the meniscus underestimates the concentration.
a. If the student read the molarity of [tex]AgNO_3[/tex] as 0.01426 M instead of 0.01246 M, the experimentally calculated moles of Ag would be too high. As a result, the calculated moles of [tex]Cl^-[/tex] would also be too high.
Since the moles of Ag and [tex]Cl^-[/tex] are in a 1:1 ratio according to the balanced chemical equation ([tex]Ag^+[/tex] + [tex]Cl^-[/tex] -> AgCl), the calculated concentration of [tex]Cl^-[/tex] would also be too high.
This error would lead to an overestimation of the chloride concentration.
b. If the student took the initial buret reading correctly but took the final buret reading from the top of the meniscus, the experimentally determined moles of Ag would be too low.
Consequently, the calculated moles of [tex]Cl^-[/tex] would also be too low. Again, since the moles of Ag and [tex]Cl^-[/tex] are in a 1:1 ratio, the calculated concentration of [tex]Cl^-[/tex] would be underestimated.
This error would result in an underestimated chloride concentration.
In both cases, the errors affect the calculated concentration of [tex]Cl^-[/tex]. The first error would lead to an overestimated concentration, while the second error would result in an underestimated concentration.
Accurate determination of the molarity and careful measurement of volumes are crucial for obtaining reliable concentration values.
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CHAPTER 12
The substance xenon has the following properties: normal melting point: \( 161.3 \mathrm{~K} \) normal boiling point: \( 165.0 \mathrm{~K} \) \( \begin{array}{ll}\text { triple point: } & 0.37 \mathrm
The true points about xenon are: The final state of substance is liquid, the initial state of sample is gas, and one or more phase changes will occur, hence options C, D, and E are correct.
The properties of the xenon are:
Normal melting point: 161.3 K
Normal boiling point: 165.0 K
Triple point: 0.37 atm, 152.0 K
Critical point: 57.6 atm, 289.7 K
As a result of ultimate temperature being below the boiling point, liquid is a substance's ultimate state.
It is in the gas phases because the starting temperature is greater than the boiling point. The sample's initial condition is one of gas.
Phase shifts might occur when temperatures fluctuate. There will be one or more phase transitions.
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The given question is incomplete, so the most probable complete question is,
The substance xenon has the following properties: normal melting point: 161.3 K normal boiling point: 165.0 K triple point: 0.37 atm, 152.0 K critical point: 57.6 atm, 289.7 K A sample of xenon at a pressure of 1.00 atm and a temperature of 188.9 K is cooled at constant pressure to a temperature of 163.1 K. Which of the following are true? Choose all that apply
A) The final state of the substance is a solid.
B) The liquid initially present will vaporize.
C) The final state of the substance is a liquid.
D) The sample is initially a gas.
E) One or more phase changes will occur.
Asubstance has a triple point at −245 ′
C and 225 mmHg. What is moat licoy to happen lo a sold sample of the substance as it is warmed from −35 ∘
C to −30 ∘
C at a pressure of 200mmHg ? The sold will sublime into a gas. Noting (the sold will rectain as a solid. The solid will met into a laud.
As the solid sample of the substance is warmed from -35 °C to -30 °C at a pressure of 200 mmHg, it will undergo sublimation, transforming directly from a solid to a gas phase.
The given information indicates that the substance has a triple point at -245 °C and 225 mmHg. A triple point is the temperature and pressure at which a substance can coexist in all three phases: solid, liquid, and gas. In this case, the triple point conditions are at -245 °C and 225 mmHg.
Since the temperature range from -35 °C to -30 °C falls above the triple point temperature, and the pressure of 200 mmHg is within the range of the triple point pressure, the substance will undergo sublimation. Sublimation is the process where a solid directly transforms into a gas without passing through the liquid phase.
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Belar are senctunts and predict product (10pT3) 21 3.
The product of (10pT3) 21 3 is 10003. This can be calculated by first multiplying 10pT3 by 21, which gives 210pT3. Then, multiplying 210pT3 by 3 gives 10003.
The expression (10pT3) 21 3 can be simplified as follows:
(10pT3) 21 3 = 10pT3 * 21 * 3
= 210pT3
= 10003
The first step is to multiply 10pT3 by 21. This gives 210pT3. Then, we multiply 210pT3 by 3 to get 10003.
The expression 10pT3 represents a molecule with 10 carbon atoms, 3 hydrogen atoms, and 1 phosphorus atom. The exponents in the expression indicate the number of times each atom appears in the molecule. For example, the 3 in pT3 indicates that there are 3 hydrogen atoms in the molecule.
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calculate the amount of heat needed to melt 129g of solid hexanel (C6H14) and bring it to a temperature of 3.5 degrees C. be sure you answer has a unit symbol and the correct number of significant digits
127g of hexane
The amount of heat needed to melt 129g of solid hexane and bring it to a temperature of 3.5 degrees C is approximately 15,916.47 J, with the correct unit symbol and significant digits.
To calculate the amount of heat needed to melt 129g of solid hexane (C6H14) and bring it to a temperature of 3.5 degrees C, we need to consider the heat required for both the melting process and the temperature change.
First, let's calculate the heat needed to melt 129g of hexane. The heat of fusion for hexane is 9.96 kJ/mol. To convert grams to moles, we divide by the molar mass of hexane, which is 86.18 g/mol.
Moles of hexane = 129g / 86.18 g/mol = 1.498 moles
Heat required to melt hexane = moles of hexane * heat of fusion
= 1.498 moles * 9.96 kJ/mol = 14.91 kJ
Next, let's calculate the heat required to raise the temperature of the melted hexane from 0 degrees C to 3.5 degrees C. The specific heat capacity of hexane is 2.22 J/g°C.
Heat required for temperature change = mass of hexane * specific heat capacity * temperature change
= 127g * 2.22 J/g°C * (3.5°C - 0°C) = 1006.47 J
Adding the two values together, we get the total amount of heat needed:
Total heat = heat required to melt + heat required for temperature change
= 14.91 kJ + 1006.47 J
To convert the heat to the correct unit and significant digits, we need to convert 14.91 kJ to J:
Total heat = (14.91 kJ * 1000 J/kJ) + 1006.47 J
= 14910 J + 1006.47 J
= 15916.47 J
Therefore, the amount of heat needed to melt 129g of solid hexane and bring it to a temperature of 3.5 degrees C is approximately 15,916.47 J, with the correct unit symbol and significant digits.
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Phosphorus pentachloride, PCl5PCl5, reacts with water to form phosphoric acid,H3P4,and hydrochloric acid, HCl, according to the following equation:
PCl5()+4H2()→H3P4(aq)+5HCl(aq)
What mass of PCl5PCl5 is needed to react with an excess quantity of H2H2O to produce 23.5 g of H3P4?
To produce 23.5 g of H3P4, 23.5 g of PCl5 is required, according to the 1:1 mole ratio between the two substances.
To determine the mass of PCl5 needed, we need to calculate the molar mass of H3P4 and then use stoichiometry.
1. Calculate the molar mass of H3P4:
H3P4 = (3 * molar mass of H) + (1 * molar mass of P)
= (3 * 1.0079 g/mol) + (1 * 30.9738 g/mol)
= 3.0237 g/mol + 30.9738 g/mol
= 34.9975 g/mol
2. Use stoichiometry to find the mass of PCl5:
According to the balanced equation, the mole ratio between H3P4 and PCl5 is 1:1.
Therefore, the mass of PCl5 needed is equal to the molar mass of H3P4.
Mass of PCl5 = 23.5 g of H3P4
Therefore, the mass of PCl5 needed to produce 23.5 g of H3P4 is 23.5 g.
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Using the following spectral information, propose a structure for the following aromatic compound. Explain how the information led you to your proposed structures. ¹HNMR: FTIR: GCMS: Singlet at 7.5 delta units, integration of 1 Doublet at 7.4 delta units, integration of 1 Doublet at 7.3 delta units, integration of 1 Triplet at 7.0 deha urins, hegration th Singlet at 2.5 delta units, integration of 3 Quartet at 2.3 delta units, integration of 2 Triplet at 1.0 delta units, integration of 3 Peaks at 3100 to 3000 cm-1, 3000 to 2800 cm-¹, 1600 to 1500 cm-¹, and 1500 to 1400 cm-1 Molecular ion peak at 120 m/z Explanation: 2. Using all the necessary resonance structures and explanations, please explain why benzoic acid is a deactivator and directs meta, while toluene is an activator and directs ortho/para.
A- Based on the spectral information, the proposed structure for the aromatic compound is 1,3,5-trimethylbenzene,
B- benzoic acid is a deactivator that directs meta while toluene is an activator that directs ortho/para due to their respective electron-withdrawing and electron-donating groups.
A- The ¹H NMR spectrum shows singlet peaks at 7.5 and 2.5 delta units, which are characteristic of aromatic protons and methyl protons, respectively. The presence of three methyl groups is supported by the integration value of 3 for the singlet peak at 2.5 delta units. Additionally, the doublet peaks at 7.4 and 7.3 delta units suggest the presence of neighboring aromatic protons.
The FTIR spectrum indicates the presence of peaks in the range of 3100 to 3000 cm⁻¹, which are characteristic of aromatic C-H stretching vibrations. The presence of peaks in the ranges of 3000 to 2800 cm⁻¹, 1600 to 1500 cm⁻¹, and 1500 to 1400 cm⁻¹ indicates the presence of aromatic C-H bending vibrations and aromatic C=C stretching vibrations.
The GC-MS data shows a molecular ion peak at 120 m/z, suggesting a molecular formula consistent with 1,3,5-trimethylbenzene.
B- Regarding the explanation of the second question, benzoic acid is a deactivator due to the presence of the electron-withdrawing carboxyl group (-COOH), which withdraws electron density from the benzene ring through resonance. This deactivating effect makes the benzene ring less reactive towards electrophilic substitution, and it directs incoming groups to the meta position, away from the carboxyl group.
On the other hand, toluene is an activator due to the presence of the electron-donating methyl group (-CH₃), which donates electron density to the benzene ring through induction. This activating effect makes the benzene ring more reactive towards electrophilic substitution, and it directs incoming groups to the ortho and para positions, adjacent or opposite to the methyl group.
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If a pound of sand contains 4.10 ∗
10 4
grains of sand, what would 1 mole of sand grains weigh (in pounds)? (select the best answer) 6.02 ∗
10 23
pounds 6.81 ∗
10 −20
pounds 6.81 ∗
10 20
pounds 1.47 ∗
10 19
pounds
If a pound of sand contains 4.10 × 10⁴ grains of sand, so the 1 mole of sand grains weigh 1.47 × 10¹⁹ pounds, hence option D is correct.
A mole is 6.02214076 × 10²³ of any chemical unit, including atoms, molecules, ions, and others. Due to the large number of atoms, molecules, or other components that make up any material, the mole is a useful measure to utilize.
By using dimensional analysis:
[tex]\rm6.022\times 10^2^3 SandGrains \times\frac{1lb}{ 4.10 \times 10^4 SandGrains }[/tex]
= 1.47 × 10¹⁹ pounds
Thus, if a pound of sand contains 4.10 × 10⁴ grains of sand, so the 1 mole of sand grains weigh 1.47 × 10¹⁹ pounds.
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1.
2. - I posted number 2 separately but they did it incorrectly.
if you could help me with this one also, that would be really
appreciated.
Draw the neutral organic product formed by the reaction of hydroxylamine hydrochloride with acetone.
Draw the three components needed to synthesize the given molecule via a Mannich reaction.
The reaction of hydroxylamine hydrochloride with acetone yields N-hydroxyacetone. To synthesize the given molecule via a Mannich reaction, a primary or secondary amine, an aldehyde or ketone, and an acid catalyst are required to form β-amino carbonyl compounds.
The reaction of hydroxylamine hydrochloride with acetone leads to the formation of a neutral organic product known as N-hydroxyacetone.
The reaction involves the nucleophilic attack of hydroxylamine ([tex]NH_2OH[/tex]) on the carbonyl carbon of acetone, followed by proton transfer and elimination of the chloride ion.
The resulting product, N-hydroxyacetone, contains a hydroxylamine functional group (-NHOH) attached to the carbonyl carbon of acetone.
Now, for the synthesis of the given molecule via a Mannich reaction, three components are required: a primary or secondary amine, an aldehyde or ketone, and a compound that acts as an acid catalyst.
The Mannich reaction is a condensation reaction that leads to the formation of β-amino carbonyl compounds.
The three components needed for the synthesis are:
1. Primary or secondary amine: This can be an amine such as methylamine ([tex]CH_3NH_2[/tex]) or dimethylamine [tex](CH_3)_2NH[/tex].
2. Aldehyde or ketone: An example could be formaldehyde (HCHO) or acetone [[tex](CH_3)_2CO[/tex]].
3. Acid catalyst: A common acid catalyst used in the Mannich reaction is para-toluenesulfonic acid (p-TsOH).
In the presence of the acid catalyst, the amine and the aldehyde/ketone undergo a condensation reaction, resulting in the formation of an iminium ion intermediate.
This intermediate subsequently undergoes nucleophilic attack by the enolate formed from the carbonyl compound, leading to the final β-amino carbonyl product.
It's important to note that the specific choices of amine, aldehyde/ketone, and acid catalyst can vary depending on the desired product and reaction conditions.
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Use the References to access important values if needed for this question. A student determines the value of the equilibrium constant to be 1.86×107 for the following reaction. Fe(s)+2HCl(aq)→FeCl2(s)+H2(g) Based on this value of Keq : ΔG∘ for this reaction is expected to be than zero. Calculate the free energy change for the reaction of 1.92 moles of Fe(s) at standard conditions at 298 K. ΔGron ∘= k]
The free energy change for the reaction of 1.92 moles of Fe(s) at standard conditions at 298 K is -39.41 kJ/mol.
For calculate the free energy change (ΔG∘) for the reaction of 1.92 moles of Fe(s) at standard conditions (298 K) based on the equilibrium constant (Keq) value of 1.86×10^7, we can use the following equation:
ΔG° = -RT ln(Keq)
where R is the gas constant (8.314 J/(molK)) and T is the temperature in Kelvin.
Plugging in the values:
ΔG° = -(8.314 J/(mol·K)) * 298 K * ln(1.86×10^7)
Calculating this expression will give us the value of ΔG° . Please note that the ln function represents the natural logarithm.
Using the given values, the calculation is as follows:
ΔG° = -(8.314 J/(mol·K)) * 298 K * ln([tex]1.86 * 10^{7}[/tex])
≈ -8.314 J/(molK) * 298 K * 16.842
Calculating this expression gives us:
ΔG° ≈ -39,410 J/mol
To convert this value to kilojoules per mole, we divide by 1000:
ΔG° ≈ -39.41 kJ/mol
Therefore, the free energy change (ΔG° ) for the reaction of 1.92 moles of Fe(s) at standard conditions at 298 K is approximately -39.41 kJ/mol.
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Based off lewis theory and lewis structures. Which compound(s)
would be expected to have both ionic and covalent bonds?
NH4ClO4
HC2H3O2
Mg(OH)2
H2SO4
H3O+
Among the compounds listed, H2SO4 is expected to have both ionic and covalent bonds.
H2SO4, also known as sulfuric acid, is a compound that can exhibit both ionic and covalent bonding. It contains both hydrogen (H) and sulfur (S) atoms. In sulfuric acid, the bonding between hydrogen and sulfur is covalent since they share electrons. However, when sulfuric acid dissolves in water, it ionizes into H+ ions and HSO4- ions. The H+ ions indicate the presence of ionic bonding, where hydrogen loses an electron to become a positively charged ion.
Ionic bonds are formed between atoms with significantly different electronegativities, resulting in the transfer of electrons from one atom to another. Covalent bonds, on the other hand, involve the sharing of electrons between atoms with similar electronegativities. In the case of H2SO4, the covalent bonding occurs within the molecule, while the ionic bonding occurs when it dissociates in water.
The other compounds listed do not exhibit both ionic and covalent bonding. NH4ClO4 (ammonium perchlorate) and Mg(OH)2 (magnesium hydroxide) have predominantly ionic bonds, while HC2H3O2 (acetic acid) and H3O+ (hydronium ion) have predominantly covalent bonds.
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A chromatographic analysis for Sample Y gives a peak with a retention time of 8.5 min and W1/2 of 0.17 min. How many theoretical plates are involved in the separation. Given that the column used in this analysis is 2.0 meters long, what is the height of the theoretical plate?
The number of theoretical plates involved in the separation is 40,000, and the height of the theoretical plate is 0.05 mm.
To calculate the number of theoretical plates involved in the separation, we can use the formula:
[tex]\[N = 16 \times \left(\frac{tR}{W_{1/2}}\right)^2\][/tex]
where:
N is the number of theoretical plates,
tR is the retention time of the peak, and
W1/2 is the peak width at half the peak height.
Given that the retention time (tR) is 8.5 min and the peak width at half height (W1/2) is 0.17 min, we can substitute these values into the formula:
N = 16 * (8.5 min / 0.17 min)²
N = 16 * (50)²
N = 16 * 2500
N = 40,000
The number of theoretical plates involved in the separation is 40,000.
To calculate the height of the theoretical plate (H), we can use the formula:
H = L / N
where:
H is the height of the theoretical plate, and
L is the length of the column.
Given that the length of the column (L) is 2.0 meters, we can substitute these values into the formula:
H = 2.0 m / 40,000
H = 0.00005 meters
The height of the theoretical plate is 0.00005 meters or 0.05 mm.
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the half-life of caesium-137 is about 30 years. how long will it take a sample of caesium-137 to decay to 33% of the original amount? round your answer to the nearest year.
The half-life of caesium-137 is about 30 years. it will take 60 years a sample of caesium-137 to decay to 33% of the original amount.
To determine the time it takes for a sample of cesium-137 to decay to 33% of the original amount, we can use the concept of half-life.
The half-life of cesium-137 is given as 30 years. This means that every 30 years, the amount of cesium-137 in the sample will reduce by half.
To find the time it takes to decay to 33% of the original amount, we need to determine the number of half-lives required.
Let's represent the original amount of cesium-137 as 100%. We want to find the time it takes for the amount to reach 33%.
33% of the original amount is equal to 33% of 100%, which is 33%.
Since each half-life reduces the amount by half, we can calculate the number of half-lives required to reach 33% by solving the equation:
(1/2)ⁿ = 0.33
where n represents the number of half-lives.
Taking the logarithm of both sides, we have:
n * log(1/2) = log(0.33)
Using the properties of logarithms, we can simplify this equation to:
n = log(0.33) / log(1/2)
n = 1.649
Since we cannot have a fraction of a half-life, we round up to the nearest whole number.
Therefore, it would take approximately 2 half-lives for the sample of cesium-137 to decay to 33% of the original amount.
Since each half-life is 30 years, the total time it would take is:
Total time = Number of half-lives * Half-life time
Total time = 2 * 30 years
Total time = 60 years
Therefore, it would take approximately 60 years for the sample of cesium-137 to decay to 33% of the original amount.
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Draw the correct Lewis dot structure of 10F s
The Lewis dot structure of 10Fs cannot be accurately represented as it implies the presence of 10 fluorine atoms bonded to a single sulfur atom, which is highly unstable and unlikely.
The Lewis dot structure is a representation of the valence electrons in an atom or molecule using dots placed around the atomic symbol. However, the concept of 10Fs implies the presence of 10 fluorine atoms bonded to a single sulfur atom.
Sulfur (S) belongs to Group 16 of the periodic table and has 6 valence electrons. Fluorine (F) belongs to Group 17 and has 7 valence electrons.
The octet rule states that atoms tend to gain, lose, or share electrons to achieve a stable electron configuration with a complete outer shell of 8 electrons (except for hydrogen, which only requires 2 electrons).
In the case of sulfur, it can form multiple bonds with fluorine, but 10 fluorine atoms bonded to a single sulfur atom would result in an extremely unstable structure.
The total number of valence electrons in such a structure would be 6 (from sulfur) + 10x7 (from fluorine) = 76 electrons, which is highly unlikely for a single atom.
Therefore, it is not possible to draw a accurate Lewis dot structure for 10Fs. It is important to note that chemical compounds are typically formed by combining atoms in ratios that allow for stable structures and follow the octet rule.
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5. An accurately measured 2g sample of hydrogen peroxide (H₂O₂-34g/mol) was dissolved in a mixture of 20mL water and 20mL diluted sulfuric acid. Sample is titrated with 0.1N potassium permanganate
We identify the moles of KMnO₄ utilized in the titration (0.003 mol) and convert them to moles of H₂O₂ (0.0075 mol) using the stoichiometric ratio to determine the proportion of hydrogen peroxide. The mass of H₂O₂ that reacted is 0.255 g, and the sample's proportion is roughly 12.75%.
To calculate the percentage of hydrogen peroxide in the sample, we need to determine the amount of hydrogen peroxide reacted during the titration. From the balanced equation, we can see that the stoichiometric ratio between hydrogen peroxide and potassium permanganate is 5:2.
First, let's calculate the number of moles of potassium permanganate (KMnO₄) used in the titration:
Molarity of KMnO₄ = 0.1 N (0.1 mol/L)
Volume of KMnO₄ used = 30 mL = 0.03 L
Number of moles of KMnO₄ = Molarity x Volume = 0.1 mol/L x 0.03 L = 0.003 mol
According to the stoichiometry of the balanced equation, 5 moles of hydrogen peroxide (H₂O₂) react with 2 moles of potassium permanganate (KMnO₄).
Therefore, the number of moles of hydrogen peroxide reacted is:
Number of moles of H₂O₂ = (0.003 mol KMnO₄) x (5 mol H₂O₂ / 2 mol KMnO₄) = 0.0075 mol
The molar mass of hydrogen peroxide (H₂O₂) is 34 g/mol.
Mass of H₂O₂ reacted = Number of moles x molar mass = 0.0075 mol x 34 g/mol = 0.255 g
Now, we can calculate the percentage of hydrogen peroxide in the sample:
Percentage of H₂O₂ = (Mass of H₂O₂ / Mass of the sample) x 100
= (0.255 g / 2 g) x 100
= 12.75%
Therefore, the percentage of hydrogen peroxide in the sample is approximately 12.75%.
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Complete question :
An accurately measured 2g sample of hydrogen peroxide (H₂O₂-34g/mol) was dissolved in a mixture of 20mL water and 20mL diluted sulfuric acid. Sample is titrated with 0.1N potassium permanganate consuming 30mL to reach the endpoint. Compute for the percentage of hydrogen peroxide. 2KMnO4 + 5H₂O₂ + 3H₂SO42MnSO4 + K₂SO4 +50₂ + 8H₂O n
Determine how many moles of O2 are required to react completely
with 5.0 mol of C4H10C4H10
32.5 moles of [tex]O_{2}[/tex] are required to react completely with 5.0 mol of [tex]C_{4}H_{10}[/tex].
The balanced chemical equation for the combustion reaction between oxygen ([tex]O_{2}[/tex]) and butane ([tex]C_{4}H_{10}[/tex]) is:
2 [tex]C_{4}H_{10}[/tex] + 13 [tex]O_{2}[/tex] -> 8 [tex]CO_{2}[/tex] + 10[tex]H_{2}O[/tex]
From the balanced equation, we can see that 13 moles of [tex]O_{2}[/tex] are required to react completely with 2 moles of [tex]C_{4}H_{10}[/tex].
Therefore, to determine the number of moles of [tex]O_{2}[/tex] required to react with 5.0 mol of C4H10, we can set up a proportion:
(5.0 mol [tex]C_{4}H_{10}[/tex]) / (2 mol [tex]C_{4}H_{10}[/tex]) = (x mol [tex]O_{2}[/tex]) / (13 mol [tex]O_{2}[/tex])
Cross-multiplying and solving for x, we get:
x = (5.0 mol [tex]C_{4}H_{10}[/tex] * 13 mol [tex]O_{2}[/tex]) / (2 mol [tex]C_{4}H_{10}[/tex]) = 32.5 mol [tex]O_{2}[/tex]
Therefore, 32.5 moles of [tex]O_{2}[/tex] are required to react completely with 5.0 mol of [tex]C_{4}H_{10}[/tex].
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Describe how to prepare 50 ml of a 5% (v/v) aqueous solution of
methanol (CH3OH m.w. 32g)
For preparing a 50 ml, 5% (v/v) aqueous solution of methanol (CH3OH), measure 2.5 ml of methanol and dilute it with distilled water in a 50 ml volumetric flask.
To prepare a 50 ml, 5% (v/v) aqueous solution of methanol (CH3OH), measure 2.5 ml of methanol using a graduated cylinder or pipette. Transfer the methanol to a 50 ml volumetric flask and fill it to the mark with distilled water. By following these steps, you can successfully prepare the desired 50 ml, 5% (v/v) aqueous solution of methanol.
To prepare a 50 ml, 5% (v/v) aqueous solution of methanol (CH3OH), follow these steps:
1. Calculate the volume of methanol needed:
Volume of methanol = (desired concentration) * (desired final volume)
= 5% * 50 ml
= 0.05 * 50 ml
= 2.5 ml
2. Measure out 2.5 ml of methanol using a graduated cylinder or pipette.
3. Transfer the measured methanol into a 50 ml volumetric flask using a funnel, if necessary.
4. Fill the volumetric flask to the 50 ml mark with distilled water, ensuring the bottom of the meniscus aligns with the mark.
5. Gently swirl the volumetric flask to ensure thorough mixing of the methanol and water.
Now, you have prepared a 50 ml, 5% (v/v) aqueous solution of methanol (CH3OH).
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B. Ksp for Ca(OH)2 dissolved in added
Ca2+ solution
Trial 1
Trial 2
Molarity of HCl
0.050
0.050
Final HCl buret reading
22.00
31.70
Initial HCl buret reading
0.00
10.00
The Ksp for Ca(OH)₂ dissolved in the added Ca²⁺ solution is approximately 8.41 x 10⁻⁵ mol²/L² for Trial 1 and 8.17 x 10⁻⁵ mol²/L² for Trial 2.
To calculate the Ksp for Ca(OH)₂, we can use the concept of neutralization. The reaction between Ca(OH)₂ and HCl can be represented as follows:
Ca(OH)₂ (aq) + 2HCl (aq) -> CaCl₂ (aq) + 2H₂O (l)
In this reaction, each mole of Ca(OH)₂ reacts with 2 moles of HCl to produce 1 mole of CaCl₂ and 2 moles of water. By determining the moles of HCl used in each trial and knowing the initial molarity of HCl, we can calculate the moles of Ca(OH)₂ that reacted.
Trial 1:
Moles of HCl used = Molarity of HCl x Volume of HCl used
= 0.050 mol/L x (22.00 mL - 0.00 mL)
= 1.10 x 10⁻³ mol HCl
Trial 2:
Moles of HCl used = Molarity of HCl x Volume of HCl used
= 0.050 mol/L x (31.70 mL - 10.00 mL)
= 1.085 x 10⁻³ mol HCl
Since 1 mole of Ca(OH)₂ reacts with 2 moles of HCl, the moles of Ca(OH)₂ reacted in each trial is half of the moles of HCl used.
Trial 1:
Moles of Ca(OH)₂ reacted = 1.10 x 10⁻³ mol HCl / 2
= 5.50 x 10⁻⁴ mol Ca(OH)₂
Trial 2:
Moles of Ca(OH)₂ reacted = 1.085 x 10⁻³ mol HCl / 2
= 5.425 x 10⁻⁴ mol Ca(OH)₂
Now, we can use the moles of Ca(OH)₂ reacted to calculate the concentration of Ca²⁺ in the solution. Since CaCl₂ dissociates into Ca²⁺ and Cl⁻ in a 1:2 ratio, the concentration of Ca²⁺ is twice the concentration of Ca(OH)₂ reacted.
Trial 1:
Concentration of Ca²⁺ = (5.50 x 10⁻⁴ mol Ca(OH)₂) / (0.060 L)
= 9.17 x 10⁻³ mol/L
Trial 2:
Concentration of Ca²⁺ = (5.425 x 10⁻⁴ mol Ca(OH)₂) / (0.060 L)
= 9.04 x 10⁻³ mol/L
Finally, we can calculate the Ksp using the concentration of Ca²⁺. Since the reaction involves the dissolution of Ca(OH)₂, the Ksp expression is given by:
Ksp = [Ca²⁺]²
Trial 1:
Ksp = (9.17 x 10⁻³ mol/L)²
= 8.41 x 10⁻⁵ mol²/L²
Trial 2:
Ksp = (9.04 x 10⁻³ mol/L)²
= 8.17 x 10⁻⁵ mol
²/L²
Therefore, the Ksp for Ca(OH)₂ dissolved in the added Ca²⁺ solution is approximately 8.41 x 10⁻⁵ mol²/L² for Trial 1 and 8.17 x 10⁻⁵ mol²/L² for Trial 2.
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help please correct answer
Which alcohol isomer of \( \mathrm{C}_{7} \mathrm{H}_{13} \mathrm{O} \) is predicted to have the highest boiling point?
1-heptanol [tex](CH_3(CH_2)_5OH)[/tex] is predicted to have the highest boiling point among the alcohol isomers of [tex]C_7H_{13}O[/tex] due to its linear structure and the ability to form hydrogen bonds most effectively.
To determine which alcohol isomer of [tex]\( \mathrm{C}_7\mathrm{H}_{13}\mathrm{O} \)[/tex] is predicted to have the highest boiling point, we need to consider the intermolecular forces present in each isomer.
The boiling point of a compound is primarily determined by the strength of intermolecular forces. The main intermolecular forces that affect boiling points are hydrogen bonding, dipole-dipole interactions, and London dispersion forces.
Among the isomers of [tex]\( \mathrm{C}_7\mathrm{H}_{13}\mathrm{O} \)[/tex], the alcohol isomer with the highest boiling point would be the one capable of forming hydrogen bonds most effectively.
Hydrogen bonding occurs when a hydrogen atom is bonded to a highly electronegative atom (such as oxygen, nitrogen, or fluorine) and is attracted to a lone pair of electrons on another electronegative atom.
In this case, we need to compare the alcohol isomers and identify the one with the most hydrogen bonding potential.
The possible alcohol isomers of [tex]\( \mathrm{C}_7\mathrm{H}_{13}\mathrm{O} \)[/tex] are:
1. 1-heptanol ([tex]CH_3(CH_2)_5OH)[/tex]
2. 2-heptanol [tex](CH_3CH_2CH(OH)CH_2CH_2CH_3)[/tex]
3. 3-heptanol [tex](CH_3CH_2CH_2(OH)CH_2CH_2CH_3)[/tex]
4. 4-heptanol [tex](CH_3CH_2CH_2CH(OH)CH_2CH_3)[/tex]
5. 5-heptanol [tex](CH_3CH_2CH_2CH_2(OH)CH_2CH_3)[/tex]
6. 6-heptanol [tex](CH_3CH_2CH_2CH_2CH(OH)CH_3)[/tex]
7. 2-methyl-1-hexanol [tex](CH_3CH_2CH(CH_3)CH(OH)CH_2CH_3)[/tex]
Among these isomers, the one with the highest boiling point would be the alcohol that has the most hydrogen bonding sites.
Looking at the structures, we can see that 2-methyl-1-hexanol [tex](CH_3CH_2CH(CH_3)CH(OH)CH_2CH_3)[/tex] has an additional methyl ([tex]CH_3[/tex]) group attached to the second carbon, compared to the other isomers.
This additional methyl group creates more branching in the molecule, reducing the ability of the molecules to come closer and form hydrogen bonds effectively.
Therefore, the isomer predicted to have the highest boiling point among the given alcohol isomers of [tex]\( \mathrm{C}_7\mathrm{H}_{13}\mathrm{O} \)[/tex] would be 1-heptanol [tex](CH_3(CH_2)_5OH)[/tex], which has a linear structure and can form hydrogen bonds most effectively.
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nitric acid reacts very slowly with thiocyabte ion to produce nitrosulfanylcarbonitrile. select what would happen to the concentration of FeSCN^2+ and absorbnace in your cuvette
1- concentration of FeSCN^2+=
2- absorbance=
The concentration of FeSCN²⁺ will decrease and the absorbance will decrease as nitric acid reacts with thiocynate ion to produce nitrosulfanylcarbonitrile. This is because the reaction consumes FeSCN²⁺, which is the species that absorbs light at the wavelength of interest.
Nitric acid reacts with thiocynate ion to produce nitrosulfanylcarbonitrile. This reaction is slow, but it will eventually consume all of the thiocynate ion.
As the thiocynate ion is consumed, the concentration of FeSCN²⁺ will decrease. This is because FeSCN²⁺ is produced in a 1:1 ratio with thiocynate ion.
The decrease in the concentration of FeSCN²⁺ will lead to a decrease in the absorbance. This is because absorbance is directly proportional to the concentration of the absorbing species.
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What is the intermolecular forces of the substances below?
1. 2-methyl-2-propanol
2. 1-butanol
3. 2-butanol
1. 2-methyl-2-propanol exhibits primarily dipole-dipole interactions due to its polar hydroxyl (OH) group.
2. 1-butanol experiences both dipole-dipole interactions and hydrogen bonding as a result of its polar hydroxyl (OH) group.
3. 2-butanol shares similar intermolecular forces as 1-butanol, involving dipole-dipole interactions and hydrogen bonding due to its polar hydroxyl (OH) group.
1. 2-methyl-2-propanol: The intermolecular forces in 2-methyl-2-propanol are primarily dipole-dipole interactions. This is because the molecule has a polar hydroxyl (OH) group, resulting in a partial positive charge on the carbon atom and a partial negative charge on the oxygen atom.
2. 1-butanol: The intermolecular forces in 1-butanol include both dipole-dipole interactions and hydrogen bonding. The molecule contains a polar hydroxyl (OH) group, allowing for hydrogen bonding between the hydrogen of one molecule and the oxygen of another. In addition, there are dipole-dipole interactions due to the presence of the polar functional group.
3. 2-butanol: Similar to 1-butanol, the intermolecular forces in 2-butanol include dipole-dipole interactions and hydrogen bonding. The molecule has a polar hydroxyl (OH) group that can participate in hydrogen bonding. The dipole-dipole interactions also contribute to the intermolecular forces in this compound.
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1. Iron metal reacts with oxygen to give iron (III) oxide according to the following reaction.
4Fe +30₂
2Fe₂O,
-
a. An ordinary iron nail (assumed to be pure iron) that contains 2.8 g of iron (MM-56
g/mol) reacts in an environment where there is 1.28 g oxygen (MM-32 g/mol). Show a
calculation to determine the limiting reactant in this reaction. (3 pts)
b. How many grams of Fe,O, (MM-160 g/mol) will be formed in the reaction? (3 pts)
c. How many grams of the excess reactant remains after the reaction stops? (3 pts)
a. Oxygen is the limiting reactant.b. 4.32 g Fe2O3 are formed.c. 0.96 g of excess oxygen remains after the reaction stops.
We are given the balanced chemical equation for the reaction between iron and oxygen. The balanced chemical equation is:4Fe + 3O2 → 2Fe2O3Now, we need to calculate the number of moles of each reactant that we have. The molar mass of iron (Fe) is 56 g/mol, and the molar mass of oxygen (O2) is 32 g/mol.Number of moles of iron:2.8 g ÷ 56 g/mol = 0.05 molNumber of moles of oxygen:1.28 g ÷ 32 g/mol = 0.04 molAccording to the balanced equation, 4 moles of iron react with 3 moles of oxygen. So, the mole ratio of iron to oxygen is 4:3. To determine which reactant is limiting, we need to compare the actual mole ratio to the required mole ratio.Actual mole ratio:0.05 mol iron ÷ 0.04 mol oxygen = 1.25:1Required mole ratio:4 mol iron ÷ 3 mol oxygen = 1.33:1Since the actual mole ratio is less than the required mole ratio, we can conclude that oxygen is the limiting reactant.Now that we have identified the limiting reactant, we can use the balanced equation to determine the amount of product formed.According to the balanced equation, 4 moles of iron react to form 2 moles of Fe2O3. The molar mass of Fe2O3 is 160 g/mol.Moles of Fe2O3 formed:0.04 mol oxygen × (2 mol Fe2O3 ÷ 3 mol oxygen) = 0.027 molFe2O3Mass of Fe2O3 formed:0.027 mol × 160 g/mol = 4.32 g Fe2O3 We know that oxygen is the limiting reactant, which means that there is some iron left over after the reaction stops. To determine how much iron is left over, we first need to determine how much iron reacted. Since 4 moles of iron react with 3 moles of oxygen, the number of moles of iron that reacted is:Moles of iron reacted:0.04 mol oxygen × (4 mol iron ÷ 3 mol oxygen) = 0.053 mol ironMass of iron reacted:0.053 mol iron × 56 g/mol = 2.98 g ironTo determine how much excess oxygen is left over, we can subtract the amount of oxygen that reacted from the total amount of oxygen we started with:Mass of excess oxygen:1.28 g - (0.04 mol oxygen × 32 g/mol) = 0.96 g excess oxygenAnswer:a. Oxygen is the limiting reactant.b. 4.32 g Fe2O3 are formed.c. 0.96 g of excess oxygen remains after the reaction stops.For such more questions on Oxygen
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Consider the van der Waals equation of state. In the limit of low density, calculate the values of B₁ (T), B₂(T), and B3(T) and relate them back to the hard-sphere and square-well potentials. What do you notice about all Virial coefficients B3(T) onwards in terms of their dependence on the excluded volume?
Van der Waals equation of state: The van der Waals equation of state is given by the following equation, which is as follows;
(P + a[n/V]^2)(V - nb) = nRT
where P is the pressure a and b are constants of integration V is the volume of the gasn is the number of moleculer is the gas constant T is the temperature a and b are constant for different gases (often referred to as van der Waals constants).Low-density limit calculation: For low-density gases, the effect of the interaction between particles is weak. As a result, the molecules act as if they were far away from each other and don't influence each other at all.
For example, B2 = 0 for hard-sphere potential since there is no attraction between the particles and B3 = 0 for square-well potential due to zero volume and attraction of square-well potential. The virial coefficient B3 onward depends on the excluded volume since they are directly related to the attractive force between the molecules (excluded volume). Hence it concludes that the virial coefficients beyond B3 depends on the excluded volume since they are related to the attractive force between the molecules.
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What is polythene in chemistry, its properties, the examples of polythene, uses of polythene and how its made
Answer:
polythene is one of the most used plastic in the world
Explanation:
Properties of polythene are
1.Economical
2.Low co-efficient of friction
3.Excellence chemical resistance
4.Good impact resistance
5.Good fatigue and wear resistance
6.Resistance to many solvents
Examples of polythene
1. Beverage bottles
2.Food bottles
3.polyester clothing
4.Rope
How it's made
It is made by the reaction of multiple ethylene molecules in the presence of catalyst
