¿What are the units that make up the 3 quantities (mass, volume of a substance and density)?

Answer:

Explanation:

a. Determine the number of valence electrons for each atom in the molecule

In this case we both atoms are halogens. Therefore we will have 7 electrons for each atom.

b. Draw the Lewis Dot structure

In this case, the formula is [tex]ICl_3[/tex], so the central atom would be "I" and the "Cl" atoms would be placed around "I". See figure 1

c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?)

In this specific case, the "I" atom don't follow the octet rule. We will have an expanded octet for iodine (more than 8 electrons).

d. Show the polarity of each bond and for the molecule by drawing in the dipole +d

The negative dipole would be placed in the atom with higher electronegativity, in this case "Cl". The positive dipole would be placed in the atom with low electronegativity, in this case "I".

I hope it helps!

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.

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

Hope this Helps!!!!

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.

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.  

Answer:

2 Hydrogen One oxygen

Explanation:

Answer: Gas–liquid–solid triple point

The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at approximately 273.1575 K (0.0075 °C; 32.0135 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).

Explanation:

It represents the equilibrium between the liquid and gas phases. The point on this curve where the vapor pressure is 1 atm is the normal boiling point of the substance. The vapor-pressure curve ends at the critical point (B), which is at the critical temperature and critical pressure of the substance.

Answer:

a. the system is at equilibrium.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]H_2+Br_2\rightleftharpoons 2HBr[/tex]

Thus, the law of mass action is given by:

[tex]Keq=\frac{[HBr]^2}{[H_2][Br_2]} =57.6[/tex]

Nonetheless, for the given point of 4.67 × 10^-3M bromine gas, 2.14 × 10^−3 hydrogen gas, and 2.40 × 10^−2M hydrogen bromide gas we should compute the reaction quotient in order to know whether the direction of the reaction is to left or to right, thus:

[tex]Q=\frac{[HBr]^2}{[H_2][Br_2]} =\frac{(2.40x10^{-2})^2}{(4.67x10^{-3})(2.14x10^{-3})} \\\\Q=57.6[/tex]

Therefore, since Keq=Q, we say that the system is at equilibrium, for that reason, the answer is a.

Best regards.

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.

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.

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

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".

Good Luck, and have a great day..

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].

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

To 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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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:

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.

good luck :)

Answer:

Lithium

Explanation:

The specific geat capacity of a substance is the energy required to raise 1 unit of that substance by one degree.

Heat energy (Q) = mc∇t

Q = heat energy

M = mass of the substance

c = specific heat capacity

∇t = change in temperature of the substance

Generally, increase in the specific heat capacity will lead to a lower final temperature likewise decrease in the specific heat capacity will lead to increase the final temperature of the substance.

From the data above, we can take just two specific heat capacity and test this theory.

Assuming we have a

Mass = 25g

Heat energy applied (Q) = 1 J

Initial temperature (T1) = 10°C

Final temperature (T2) = ?

Q = mc∇t

Q = mc (T2 - T1)

For Lithium, specific heat capacity = 3.58J/g°C

1 = 25 × 3.58 (T2 - 10)

Solve for T2

1 = 89.5 (T2 - 10)

1 = 89.5T2 - 895

89.5T2 = 896

T2 = 896 / 89.5

T2 = 10.011°C

For Magnesium (Mg) specific heat capacity = 1.02J/g°C

Q = mc∇t

1 = 25 × 1.02 × (T2 - 10)

1 = 25.5 (T2 - 10)

1 = 25.5T2 - 255

Solve for T2

25.5T2 = 256

T2 = 10.039°C

Notice the trend that decrease in the specific heat capacity leads to increase in the final temperature.

Try and continue for the elements and see how it works.

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]

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

b. Boiling point:

ΔT = 0.51°C/m×3.5325m×1

ΔT = 1.80°C

Tfinal - 100°C = 1.80°C

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.

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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.

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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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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Answer:

The answer is Hg.

Explanation:

Symbol for Mercury is Hg.

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

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

Answer:

The class of this compound is aldehyde or ketone (i).

Explanation:

Absorption peak at 1720 cm-1 shows the presence of a carbonyl group, possibly an aldehyde or ketone with C=O bond.

Further information on molecular formula would be required for structural elucidation.

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

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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

Answer:

Explanation:boom

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!