Answer:
ΔH = - 44.0kJ
Explanation:
H2O(l)→H2O(g), ΔH =44.0kJ
In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH is positive.
If he reaction is reversed, we have;
H2O(g)→H2O(l)
In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still have the same value.
g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Answer:
The concentration of the sodium and arsenate ions at the end of the reaction in the final solution
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Explanation:
Complete Question
A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)
Concentration in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = (Concentration in mol/L) × (Number of moles)
For Na₂HAsO₄
Concentration in mol/L = 0.03798 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03798 × 0.5 = 0.01899 mole
For NaOH
Concentration in mol/L = 0.03428 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03428 × 0.5 = 0.01714 mole
Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
0.01899 0.01714 0 0 (At time t=0)
(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)
0.00185 | 0 → 0.01714 0.01714 (end)
Hence, at the end of the reaction, the following compounds have the following number of moles
Na₂HAsO₄ = 0.00185 mole
This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction
NaOH = 0 mole
Na₃AsO₄ = 0.01714 moles
This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction
H₂O = 0.01714 moles
So, at the end of the reaction
Na⁺ has 0.0037 + 0.05142 = 0.05512 mole
(HAsO₄)²⁻ has 0.00185 mole
(AsO₄)³⁻ has 0.01714 mole.
And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L
Hence, the concentration of the sodium and arsenate ions at the end of the reaction is
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Hope this Helps!!!
Answer:
[tex]\rm [Na^{+}]= \text{0.055 12 mol/L}[/tex]
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \text{0.018 99 mol/L}[/tex]
Explanation:
The overall equation for the reaction is
Na₂HAsO₄ + NaOH ⟶ Na₃AsO₄ + H₂O
1. Mass balance for Na
All the Na⁺ comes from the Na₂HAsO₄ and the NaOH.
The mass balance equation for Na is
[tex]\rm c_{Na^{+}} = 2[Na^{+}]_{Na_{2}HAsO_{4}} + [Na^{+}]_{NaOH}[/tex]
At the moment of mixing and before the reaction started, the total volume had doubled, so the concentrations of each component were halved.
[Na₂HAsO₄] = ½ × 0.037 98 =0.018 99 mol·L⁻¹
[NaOH] = ½ × 0.034 28 = 0.017 14 mol·L⁻¹
[tex]\rm c_{Na^{+}} = 2\times 0.01899 + 0.01714 = 0.03798 + 0.01714\\c_{Na^{+}}= \textbf{0.055 12 mol/L}[/tex]
2. Mass balance for arsenate species
All the arsenate species come from the Na₂HAsO₄.
The reactions involved are
HAsO₄²⁻+ OH⁻ ⇌ AsO₄³⁻ + H₂O
HAsO₄²⁻ + H₂O ⇌ H₂AsO₄⁻ + OH⁻
H₂AsO₄⁻ + H₂O ⇌ H₃AsO₄ + OH⁻
The mass balance equation for arsenate species is
[tex]\rm c_{\text{arsenate}} = [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}][/tex]
At the moment of mixing, the concentration of Na₂HAsO₄ had halved.
[Na₂HAsO₄] = ½ × 0.039 78 = 0.018 99 mol·L⁻¹
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \textbf{0.018 99 mol/L}[/tex]
what is a mitochondrion
Explanation:
Mitochondria (sing. mitochondria) are organelles, or parts of the eukaryote cell. They are in the cytoplasm, not the nucleus. They make the most cell supply of adenosine triphosphate (ATP), a molecule that cells use as an energy source. ... This means that mitochondria are known as '' the powerhouse of the cell'' or ''cell strength".
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a) A molecule of DNA contains 140 A bases, how many T bases will it contain? Plz help ❤️
Answer:
140 T or thymine base
Explanation:
Adenine pairs with Thymine in DNA thus the number of adenine will always equal number of thymine (unless some sort of mutation), therefore in this problem you have 140 A so you have 140 T as well. Remember: Adenine (A) and Thymine(T) is equal, & Cytosine (C) and Guanine (G) is equal
A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur. Solution A: 3.75 % (m/v) NaCl Solution B: 1.92 % (m/v) glucose Solution C: distilled H2O Solution D: 9.03 % (m/v) glucose Solution E: 5.0% (m/v) glucose and 0.9% (m/v) NaCl
Answer:
A. The cell will undergo crenation
B. The cell will undergo hemolysis
C. The cell will undergo hemolysis
D. The cell will undergo crenation.
E. The cell will undergo neither crenation nor hemolysis
Explanation:
A hypotonic solution is a solution in which the concentration of solutes is greater inside the cell than outside the cell.
A hypertonic solution is a solution in which the concentration of solutes is greater outside the cell than inside the cell.
An isotonic solution is a solution in which the concentration of solution is the same outside and inside of the cell. A solution with 5% (m/v) glucose and 0.9% (m/v) NaCl is an isotonic solution.
When a red blood cell is placed in a hypotonic solution, it will swell and burst. This is known as hemolysis.
When placed in a hypertonic solution, a red blood cell will lose water and shrivel. This is is known as cremation.
When a red blood cell is placed in an isotonic solution, neither hemolysis or crenation occurs as there is no net movement of water across the cell's membrane.
A: 3.75 % (m/v) NaCl Solution is a hypertonic solution. The cell will undergo crenation
B: 1.92 % (m/v) glucose Solution is hypotonic. The cell will undergo hemolysis
C: Distilled H2O Solution is a hypotonic solution. The cell will undergo hemolysis
D: 9.03 % (m/v) glucose Solution is a hypertonic solution. The cell will undergo crenation
E: 5.0% (m/v) glucose and 0.9% (m/v) NaCl are both isotonic solutions. The cell will undergo neither hemolysis not crenation.
Solution A (3.75% NaCl): Crenation
Solution B (1.92% glucose): HemolysisSolution C (Distilled H2O): HemolysisSolution D (9.03% glucose): CrenationSolution E (5.0% glucose and 0.9% NaCl): CrenationWhat is the SolutionTo find out if a red blood cell will shrink, burst, or stay the same when placed in a solution, we have to think about how concentrated the solution is compared to the red blood cell.
A red blood cell has a normal concentration of about 0. 9% salt or 0. 3% sugar Solutions that have more concentrated substances are called hypertonic, while solutions that have less concentrated substances are called hypotonic.
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Amylase is the enzyme that controls the breakdown of starch to glucose. Describe how the student could investigate the effect of pH on the breakdown of starch by amylase.
Answer:
Explanation:
You will investigate the breakdown of starch by amylase at different pHs.
The different pHs under investigation will be produced using buffer solutions. Buffer solutions produce a particular pH, and will maintain it if other substances are added.
The amylase will break down the starch.
A series of test tubes containing a mixture of starch and amylase is set up at different pHs.
A sample is removed from the test tubes every 10 seconds to test for the presence of starch. Iodine solution will turn a blue/black colour when starch is present, so when all the starch is broken down, a blue-black colour is no longer produced. The iodine solution will remain orange-brown.
A control experiment must be set up - without the amylase - to make sure that the starch would not break down anyway, in the absence of an enzyme. The result of the control experiment must be negative - the colour must remain blue-black - for results with the enzyme to be valid.
When the starch solution is added:
Start timing immediately.Remove a sample immediately and test it with iodine solution.Sample the starch-amylase mixture continuously, for example every 10 seconds.For each pH investigated, record the time taken for the disappearance of starch, ie when the iodine solution in the spotting tile remains orange-brown.
The time taken for the disappearance of starch is not the rate of reaction.
It will give us an indication of the rate, but is the inverse of the rate - the shorter the time taken, the greater the rate of the reaction.
We can calculate the rate of the reaction by calculating \frac{1}{t}, obtaining a measure of the rate of reaction by dividing one by the time taken for the reaction to occur.
A similar experiment can be carried out to investigate the effect of temperature on amylase activity.
Set up a series of test tubes in the same way and maintain these at different temperatures using a water bath - either electrical or a heated beaker of water.
Depending on the chemical reaction under investigation, you might monitor the reaction in a different way. If investigating the effect of temperature on the breakdown of lipid by lipase, you could monitor pH change - lipids are broken down into fatty acids and glycerol. As the reaction begins, the release of fatty acids will mean that the pH will decrease.
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Which of the following best describe an atomic number? An element identity is defined by its atomic number this means it represents the number of
a)Atomic number is the number of protons in the nucleus of an atom.
b)atomic number represents the number of protons
An element's identity is defined by its atomic number; this means it represents the number of protons in its nucleus.
The question is incomplete, the complete question is;
Which of the following best describes an atomic number? An element's identity is defined by its atomic number; this means it represents the number of
A) protons plus neutrons in its nucleus.
B) electrons in the element.
C) protons in its nucleus.
D) neutrons in its nucleus.
The subatomic particles in the atom are;
ElectronsProtonsNeutronsNeutrons and protons are contained in the nucleus hence they are collectively called nucleons.
Electrons are found in the orbits. The number of protons must be equal to the number of electrons for the atom to be electrically neutral.
The number of protons in an atom or atomic number serve as a means of identifying an atom.
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Magnesium and nitrogen react in a combination reaction to produce magnesium nitride:
3 Mg + N2 → Mg3N2
In a particular experiment, a 8.33-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.
Answer:
21.7 g
Explanation:
Step 1: Write the balanced equation
3 Mg + N₂ → Mg₃N₂
Step 2: Calculate the moles corresponding to 8.33 g of nitrogen
The molar mass of N₂ is 28.01 g/mol.
[tex]8.33 g \times \frac{1mol}{28.01g} =0.297mol[/tex]
Step 3: Calculate the moles of magnesium that reacts with 0.297 moles of nitrogen
The molar ratio of Mg to N₂ is 3:1. The reacting moles of Mg are 3/1 × 0.297 mol = 0.891 mol
Step 4: Calculate the mass corresponding to 0.891 moles of magnesium
The molar mass of Mg is 24.31 g/mol.
[tex]0.891 mol \times \frac{24.31g}{mol} = 21.7 g[/tex]
Answer:
[tex]m_{Mg}=21.7 g Mg[/tex]
Explanation:
Hello,
In this case, considering the given reaction, we are able to compute the mass of magnesium that is consumed by considering its molar mass (24.31 g/mol), the molar mass of diatomic nitrogen (28.02 g/mol), the initial mass of nitrogen (8.33 g) and the 3:1 molar ratio of magnesium to nitrogen in the reaction.
Hence we compute it by applying the shown below stoichiometric procedure:
[tex]m_{Mg}=8.33 gN_2*\frac{1molN_2}{28.02gN_2} *\frac{3molMg}{1molN_2} *\frac{24.31gMg}{1molMg} \\\\m_{Mg}=21.7 g Mg[/tex]
Regards.
A scientist mixed 25.00 mL of 2.00 M KOH with 25.00 mL of 2.00 M HBr. The temperature of the mixed solution rose from 22.7 oC to 31.9 oC. Calculate the enthalpy change for the reaction in kJ/mol HBr, assuming that the calorimeter loses negligible heat, that the volumes are additive, and that the solution density is 1.00 g/mL, and that its specific heat is 4.184 J/g.oC.
Answer:
38.493 KJ/mol
Explanation:
Equation of reaction; HBr + KOH ---> KBr + H2O
Heat evolved = mass * specific heat capacity * temperature rise
Mass of solution = density * volume
Mass = 1.00 g/ml*50 ml = 50g
Temperature rise = 31.9 - 22.7 = 9.2 °C
Heat evolved = 50 * 4.184 * 9.2 = 1924.64 J
From the equation of reaction, 1 mole of HBr reacts with 1mole of KOH to produce 1 mole of H20
Number of moles of HBr involved in the reaction = molar concentration * volume (L)
Molar concentration = 2.0 M, volume = 25 ml = 0.025 L
Number of moles = 2.0 M * 0.025 L= 0.05 moles
Therefore, 0.05 moles of HBr reacts with 0.05 moles of KOH to produce 0.05 moles of H20
Enthalpy change per mole of HBr = 1924.64 J/0.05 moles = 38492.8 J/mol = 38.493 KJ/mol
Reactive oxygen species (ROS) are unstable, or reactive, compounds that result from the partial reduction of oxygen. ROS can cause damage to molecules, including membrane lipids and nucleic acids, and may be associated with some diseases. Which of these compounds are reactive oxygen species? Choose all that apply.
a. OH
b. OH-
c. O2-
d. H2O
e. H2O2
f. H-
4-Methylphenol, CH3C6H4OH (pKa 10.26), is only slightly soluble in water, but its sodium salt, CH3C6H4O-Na , is quite soluble in water. Describe the solubility of 4-methylphenol in solutions of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3). The pKa values for the conjugate acids of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3) are 15.7, 6.36, and 10.33, respectively.
Explanation:
We know that more is the [tex]pK_{a}[/tex] value, weaker will be the acid. Also, an acid completely dissociates into ions in an aqueous base solution when [tex]pK_{a}[/tex] of conjugate acid of base is greater than acid.
4-methylphenol [tex](CH_{3}C_{6}H_{4}OH)[/tex] ([tex]pK_{a} = 10.26[/tex]) is quite soluble in its sodium salt. In NaOH, the dissociation will be as [tex]Na^{+}[/tex] and [tex]OH^{-}[/tex] ions as NaOH is a strong base.
Therefore, 4-methylphenol will readily dissolve in NaOH solution.
As, [tex]NaHCO_{3}[/tex] is not a strong base but as 4-Methylphenol forms a sodium salt hence, it will have a low solubility as compared to NaOH.
Whereas [tex]Na_{2}CO_{3}[/tex] is not a base but when dissolved in water it shows basic character as it produces NaOH (strong base) and [tex]H_{2}CO_{3}[/tex] (weak acid). As a result, the solution gets basic. Hence, 4-methylphenol will readily dissolve in [tex]Na_{2}CO_{3}[/tex].
A 20.0-mL sample of lake water was acidified with nitric acid and treated with excess KSCN to form a red complex (KSCN itself is colorless). The solution was then diluted to 50.0-mL and put in a 1.00 cm pathlength cell, where it yielded an absorbance of 0.345. For comparison, a 5.0-mL reference sample of 4.80 x 10-4 M Fe3 was treated with HNO3 and KSCN and diluted to 50.0 mL. The reference solution was also placed in a 1.00-cm cell and gave an absorbance of 0.512. What is the concentration of Fe3 in Jordan Lake
Answer:
8.09x10⁻⁵M of Fe³⁺
Explanation:
Using Lambert-Beer law, the absorbance of a sample is proportional to its concentration.
In the problem, the Fe³⁺ is reacting with KSCN to produce Fe(SCN)₃ -The red complex-
The concentration of Fe³⁺ in the reference sample is:
4.80x10⁻⁴M Fe³⁺ × (5.0mL / 50.0mL) = 4.80x10⁻⁵M Fe³⁺
Because reference sample was diluted from 5.0mL to 50.0mL.
That means a solution of 4.80x10⁻⁵M Fe³⁺ gives an absorbance of 0.512
Now, as the sample of the lake gives an absorbance of 0.345, its concentration is:
0.345 × (4.80x10⁻⁵M Fe³⁺ / 0.512) = 3.23x10⁻⁵M.
As the solution was diluted from 20.0mL to 50.0mL, the concentration of Fe³⁺ in Jordan lake is:
3.23x10⁻⁵M Fe³⁺ × (50.0mL / 20.0mL) = 8.09x10⁻⁵M of Fe³⁺
The concentration of Fe³⁺ in Jordan Lake is = 8.09* 10⁻⁵ M
According to Lambert-Beer law ;The rate of absorbance of a sample is directly proportional to concentration of the sample
The reaction that produces a red complex
Fe³⁺ + KScN ----> Fe ( SCN )₃ ( red complex )
First step: Determine the Concentration of Fe³⁺ in reference sample
= 4.80x10⁻⁴ * ( 5.0 / 50.0 ) = 4.80 * 10⁻⁵M
reference sample was diluted from 5.0 mL to 50.0 mL
∴ Concentration of 4.80 * 10⁻⁵M has an absorbance = 0.512
Given that Lake sample absorbance = 0.345
Next step : Determine the concentration of the lake sample
Concentration of lake sample :
= absorbance of lake sample * ( conc of reference sample / absorbance )
= 0.345 * ( 4.80* 10⁻⁵ / 0.512 ) = 3.23* 10⁻⁵M.
Final step : Determine the concentration of Fe³⁺ in Jordan lake
= 3.23 * 10⁻⁵ * ( 50.0mL / 20.0mL) = 8.09* 10⁻⁵ M
Note : Solution was diluted from 20.0 mL to 50.0 mL
Hence we can conclude that The concentration of Fe³⁺ in Jordan Lake is = 8.09* 10⁻⁵ M .
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what is the sign of Mercury
Answer:
The answer is Hg.
Explanation:
Symbol for Mercury is Hg.
why does Portland have moderate climate
Answer:
Due to its location, Portland has a moderate climate. Also, which portland are you talking about? Portland Oregon or Portland Maine. Portland Oregon has a much more rainy climate, so I would say Maine has a better climate.
Explanation:
Gold can be separated from sand by panning or by a sluice-box. In panning, water is mixed with the sand and the resulting slurry is swirled in a shallow, saucer-shaped metal pan. In the sluice-box technique, running water is passed over an agitated sand- gold mixture. What physical property and what technique make this separation possible?
Answer:
Decantation by means of difference in relative densities.
Explanation:
The specific gravity (relative density) of the gold to the soil/sand is the physical property exploited in panning gold. The particles with lower density would float whilst the heavier gold sinks lower to the bottom of the pan by gravity and is decanted off.
I hope this explanation is easy to comprehend.
A sample of carbon dioxide gas at a pressure of 879 mm Hg and a temperature of 65°C, occupies a volume of 14.2 liters. Of the gas is cooled at constant pressure to a temperature of 23°C, the volume of the gas sample will be
Answer:
The correct answer is 12.43 Liters.
Explanation:
Based on the given question, the volume V₁ occupied by the sample of carbon dioxide gas is 14.2 liters at temperature (T₁) 65 degree C or 65+273 K = 338 K.
The gas is cooled at a temperature (T₂) 23 degree C or 273+23 K = 296 K
The volume of the gas (V₂) after cooling can be determined by using the formula,
V₁/T₁ = V₂/T₂
14.2/338 = V₂/296
0.0420 = V₂/296
V₂ = 0.0420 * 296
V₂ = 12.43 Liters.
1. Reaccionan 9.7 Kg de un mineral de níquel al 70% con 8L de una solución de ácido fosfórico al 60% y con una densidad de 1.36g/ml.
Answer:
The reaction produces 201.4 g of hydrogen gas and 12.2 kg of Nickel Phosphate.
Explanation:
English Translation
9.7 Kg of a 70% nickel mineral react with 8L of a 60% phosphoric acid solution and with a density of 1.36g / ml.
Solution
The problem doesn't seen to be complete as it doesn't ask a question in the end. But, we will just calculate the amount of each product expected to cover the grounds.
The balanced chemical reaction between Nickel and Phosphoric acid is given as
3Ni + 2H₃PO₄ → 3H₂ + Ni₃(PO₄)₂
We need to first obtain the limiting reagent, that is, the reagent that is used up during the reaction and is in short supply. This reagent determines the amount of products that will be formed.
Mass of nickel that is present at the start = 70% of 9.7 kg = 6.79 kg
Mass of Phosphoric acid present at the start of the reaction = 60% of (8000 mL × 1.36 g/mL) = 6528 g = 6.528 kg
Converting both of these to number of moles
Number of moles = (mass)/(Molar mass)
For nickel,
Mass = 6.79 kg = 6790 g
Molar mass = 58.6934 g/mol
Number of moles at the start = (6790/58.6934) = 115.7 moles
For Phosphoric acid
Mass = 6528 g
Molar mass = 97.994 g/mol
Number of moles = (6528/97.994) = 66.6 moles
3 moles of Ni reacts with 2 moles of H₃PO₄
From the number of moles present initially, shows that Phosphoric acid is in limited supply and is the limiting reagent.
From the stoichiometric balance of the reaction
2 moles of H₃PO₄ gives 3 moles of H₂
66.6 moles of H₃PO₄ will give (66.6×3/2) of H₂, that is, 99.9 moles of H₂.
Mass of H₂ liberated from the reaction = (Number of moles) × (molar mass) = 99.9 × 2.016 = 201.3984 g = 201.4 g
2 moles of H₃PO₄ gives 1 mole of Ni₃(PO₄)₂
66.6 moles of H₃PO₄ will give (66.6×1/2) of Ni₃(PO₄)₂, that is, 33.3 moles of Ni₃(PO₄)₂.
Mass of Ni₃(PO₄)₂ produced from the reaction = (Number of moles) × (molar mass) = 33.3 × 366.02 = 12,188.466 g = 12.2 kg
Hope this Helps!!!
An unknown compound, B, has the molecular formula C7H12. On catalytic hydrogenation 1 mol of B absorbs 2 mol of hydrogen and yields 2-methylhexane. B has significant IR absorption band at about 3300 and 2200 cm-1. Which compound best represents B?
Answer:
5-methylhex-1-yne//5-methylhex-2-yne//2-methylhex-3-yne
Explanation:
We have to start with the information on the IR spectrum. The signal at 3300 is due to a C-H bend sp carbon and the signal in 2200 is due to the stretch carbon-carbon. Therefore we will have an alkyne. Now if we have 2-methylhexane as the product of hydrogenation we have several options to put the triple bond. Between carbons 1 and 2 (5-methylhex-1-yne), between carbons 2 and 3 (5-methylhex-2-yne) and between carbons 3 and 4 (2-methylhex-3-yne). On carbon 5 we have a tertiary carbon therefore we dont have any other options.
See figure 1
I hope it helps!
A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 44./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 24.01g water 13.10g Use this information to find the molecular formula of X.
Answer:
C3H6.
Explanation:
Data obtained from the question:
Mass of the compound = 8g
Mass of CO2 = 24.01g
Mass of H2O = 13.10g
Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 44g/mol
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01
=> 12/44 x 24.01 = 6.5g
Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1
=> 2x1/18 x 13.1 = 1.5g
Mass of O in the compound = Mass of compound – (mass of C + Mass of H)
=> 8 – (6.5 + 1.5) = 0
Next, we shall determine the empirical formula of the compound. This is illustrated below:
C = 6.5g
H = 1.
Divide by their molar mass
C = 6.5/12 = 0.54
H = 1.4/1 = 1.
Divide by the smallest
C = 0.54/0.54 = 1
H = 1/0.54 = 2
Therefore, the empirical formula is CH2
Finally, we shall determine the molecular formula as follow:
The molecular formula of a compound is a multiple of the empirical formula.
Molecular formula = [CH2]n
[CH2]n = 44
[12 + (2x1)]n = 44
14n = 44
Divide both side by 14
n = 44/14
n = 3
Molecular formula = [CH2]n = [CH2]3 = C3H6
Therefore, the molecular formula of the compound is C3H6
When vinylcyclohexane is treated with in dichloromethane, the major product is (2-bromo ethylidene)cyclohexane . Account for the formation of this product by drawing the structure of the most stable radical intermediate. Include all valence lone pairs in your answer. Include all valence radical electrons in your answer.
Answer:
Explanation:
Vinylcyclohexane is an example of a cyclic hydrocarbon where the vinyl group (-CH=CH₂ ) attaches itself to an end of a cyclohexane in ring form thereby giving rise to a vinylcyclohexane. The vinyl group are ethylene with a reduction in one hydrogen atom given them the name vinyl.
SOo, when vinylcyclohexane is treated with NBS ( i.e N-Bromosuccinimide a chemical reagent used in organic reactions) ; the bromine in the NBS reacts with the cyclohexane thereby giving rise to a allyl radical first. The allyl radical is resonance stabilized radical with an unpaired electron on the allylic carbon . As a result of stabilization ; a more stable substituted cycloalkene is formed as an intermediate .
This stable substituted cycloalkene intermediate then finally react with a bromine ion to give a major product known as ; (2-bromo ethylidene)cyclohexane.
The diagram emphasizing more on the above explanation can be seen in the attached image below
An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water.
Required:
a. Determine the freezing point of the solution. Express you answer in degrees Celsius. (Assume a density of 1.00 g/mL for water.)
b. Compute the boiling point of the solution. (Assume a density of 1.00 g/mL for water.)
Answer:
a. TFinal = -6.57°C
b. Tfinal = 101.80°C
Explanation:
When a solute is added to a solvent producing an ideal solution, the freezing point of the solution decreases with regard to pure solvent. Also, boiling point increases with regard to pure solvent.
The formulas are:
Freezing point:
ΔT = Kf×m×i
Where Kf is freezeing point depression constant of water (1.86°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).
Boiling point:
ΔT = Kb×m×i
Where K is freezeing point depression constant of water (0.51°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).
Moles of 21.4g of ethylene glycol (Molar mass: 62.07g/mol) are:
21.4g C₂H₆O₂ ₓ (1mol / 62.07g) = 0.345 moles
And kg of 97.6mL of water = 97.6g are 0.0976kg. Molality of the solution is:
0.345mol / 0.0976kg = 3.5325m
Replacing in the formulas:
a. Freezing point:
ΔT = 1.86C/m×3.5325m×1
ΔT = 6.57°C
0°C - Tfinal = 6.57°C
TFinal = -6.57°Cb. Boiling point:
ΔT = 0.51°C/m×3.5325m×1
ΔT = 1.80°C
Tfinal - 100°C = 1.80°C
Tfinal = 101.80°C
which statements describe how chemical formulas, such as H2O, represent compounds?
Answer:
2 Hydrogen One oxygen
Explanation:
a introduction paragraph about coal and natural gas
Answer:
here is ur and
Explanation:
Coal, oil and natural gas are called fossil fuels. Fossil fuels are burned to make energy. Burning fossil fuels also releases CO2 (carbon dioxide) gas into the atmosphere. Most air pollutants (such as sulfur dioxide) don't stay in the atmosphere very long.
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Which is true regarding a water molecule?
Answer:
Has many answers, but one is that it consists of small polar v shaped molecules with a molecular formula H20.
Explanation:
Water molecules consists of 2 hydrogen atoms bonded with on oxygen atom. Each molecule is electrically neutral but polar, with the center of positive and negative charges located in different places.
Each hydrogen atom has a nucleus consisting of a single positively-charged proton surrounded by a 'cloud' of a single negatively-charged electron and the oxygen atom has a nucleus consisting of eight positively-charged protons and eight uncharged neutrons surrounded by a 'cloud' of eight negatively-charged electrons.
Hoped this helped!
Who proposed the plum pudding model and what does it say about the structure of the atom
Answer:
J. J. Thomson
Explanation:
First proposed by J. J. Thomson in 1904 soon after the discovery of the electron, but before the discovery of the atomic nucleus, the model tried to explain two properties of atoms then known: that electrons are negatively-charged particles and that atoms have no net electric charge.
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. This reaction describes a non-Markovnikov orientation.
In the reaction between an unsymmetrical alkene (such as propene) and N-bromosuccinimide (NBS) in the presence of aqueous dimethyl sulfoxide (DMSO), the major product is formed with the bromine atom bonded to the less highly substituted carbon atom of the alkene.
In Markovnikov's addition, the major product is formed by adding the electrophile (in this case, the bromine atom) to the carbon atom with more hydrogen atoms bonded to it. However, the given reaction exhibits non-Markovnikov selectivity, as the bromine atom adds to the less substituted carbon atom.
This non-Markovnikov selectivity can be attributed to the presence of DMSO, which acts as a polar solvent and helps generate a bromine radical (Br•). The radical intermediate can then undergo reaction with the alkene, leading to the observed regioselectivity where the bromine atom adds to the less substituted carbon. This process is known as a radical addition reaction.
Hence, the reaction demonstrates a non-Markovnikov orientation due to the addition of the bromine atom to the less highly substituted carbon atom of the propene molecule.
Learn more about the Markovnikov rule here:
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Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of . Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to significant digits.
Answer:
Concentration of Copper (II) Sulfate in the original sample in mol/L = 0.0035 M
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.56 g/L
Explanation:
Complete Question
Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
Solution
Noting that the precipitate is Copper as it is the only solid by-product of this reaction.
89 mg of Copper is produced from this reaction.
We convert this into number of moles for further stoichiometric calculations
Mass of Copper = 89 mg = 0.089 g
Molar mass of Copper = 63.546 amu
Number of moles of Copper produced from the reaction = (0.089/63.546) = 0.0014005602 = 0.001401 mole
From the stoichiometric balance of the reaction,
1 mole of Copper is produced from 1 mole of Copper (II) Sulfate
0.001401 mole of Copper will be produced similarly from 0.001401 mole of Copper (II) Sulfate.
Number of moles of Copper (II) Sulfate in the original sample = 0.001401 mole
Concentration of Copper (II) Sulfate in the original sample in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = 0.001401 mole
Volume in L = (400/1000) = 0.4 L
Concentration of Copper (II) Sulfate in the original sample in mol/L = (0.001401/0.4) = 0.0035025 mol/L = 0.0035 mol/L to 2 s.f.
Concentration in g/L = (Concentration in mol/L) × (Molar Mass)
Concentration in mol/L = 0.0035025 M
Molar mass of Copper (II) Sulfate = 159.609 g/mol
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.0035025 × 159.609 = 0.559 g/L = 0.56 g/L to 2 s.f
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The concentration of the original copper solution is 0.035 M.
The equation of the reaction is;
Fe(s) + CuSO4(aq) -------> FeSO4(aq) + Cu(s)
Number of moles of copper obtained = 89 × 10^-3g/63.5 = 0.0014 moles
Since the reaction is 1:1, the number of moles of copper sulfate that reacted is c.
From the question, we are told that the volume of solution is 400.mL or 0.04L.
Hence, the concentration of the solution is; number of moles /volume
= 0.0014 moles/0.04L = 0.035 M
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Missing parts;
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
How many moles of CO2 are produced when 84 0 mol O2 completely react?
Answer:
Explanation:boom
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
Answer:
0.136g
Explanation:
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
[tex]Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)[/tex]
Initial mole of Co(NO3)2 [tex]=\frac{mass}{molar mass}[/tex]
[tex]=\frac{5.00}{182.94} \\\\=0.02733mol[/tex]
Mole of Co(NO3)2 in final solution
[tex]=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol[/tex]
Mole of NO3- in final solution = 2 x Mole of Co(NO3)2
[tex]=2\times 0.001093\\\\=0.002186mol[/tex]
Mass of NO3- in final solution is mole x Molar mass of NO3
[tex]=0.002186\times62.01\\\\=0.136g[/tex]
The final solution contains 0.24 g of nitrate ion.
Number of moles of Co(NO3)2 = 5.00 g/183 g/mol = 0.027 moles
Number of moles = concentration × volume
concentration = Number of moles /volume
Volume of solution = 100 mL or 0.1 L
concentration = 0.027 moles/0.1 L = 0.27 M
Using the dilution formula;
C1V1 = C2V2
C1 = 0.27 M
V1 = 4.00 mL
C2 = ?
V2 = 275. mL
C2 = C1V1/V2
C2 = 0.27 × 4.00/ 275
C2 = 0.0039 M
Number of moles of NO3- ion in Co(NO3)2 = 0.0039 M × 62 g/mol = 0.24 g
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The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 50 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 500 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg
Answer:
Explanation:
a )
m = m₀ [tex]e^{-\lambda t[/tex]
m is mass after time t . original mass is m₀ , λ is disintegration constant
λ = .693 / half life
= .693 / 1590
= .0004358
m = m₀ [tex]e^{- 0.0004358 t}[/tex]
b )
m = 50 x [tex]e^{-.0004358\times 500}[/tex]
= 40.21 mg .
c )
40 = 50 [tex]e^{-.0004358t[/tex]
.8 = [tex]e^{-.0004358t[/tex]
[tex]e^{.0004358t[/tex] = 1.25
.0004358 t = .22314
t = 512 years .
Suppose that while you're in the lab performing a simple distillation you encountered one of the following errors: The components within the mixture, isopropanol and dichloromethane, distilled well below their boiling point. Poor separation between isopropanol and dichloromethane was observed. The initial volume of the distillation mixture has decreased significantly, almost dry, but no distillate was collected.
Answer:
Isopropanol and dichloromethane, distilled well below their boiling point.
Explanation:
The best way to separate isopropanol and dichloromethane is the method of fractional distillation. In this method, different compounds separate from each other due to difference in boiling. The boiling point of dichloromethane is 39.6 degree Celsius which is lower than the boiling point of isopropanol which is 82.5 degree Celsius. So dichloromethane will be evaporated when the temperature reaches to 40 degree Celsius and separated from isopropanol before reaching its boiling point.
