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Semmy [17]
2 years ago
7

(b)

Chemistry
1 answer:
Luda [366]2 years ago
3 0

The maximum mass of anhydrous zinc chloride that could be obtained from the products of the reaction is 8.18 g

<h3>Stoichiometry </h3>

From the question, we are to determine the maximum mass of anhydrous zinc chloride that could be obtained

From the given balanced chemical equation,

ZnO + 2HCl → ZnCl₂ + H₂O

This means 1 mole of ZnO will completely react with 2 moles of HCl to produce 1 mole of ZnCl₂ and 1 mole of H₂O

From the given information

Number of moles of ZnO = 0.0830 mole

Now, we will calculate the number of moles of HCl that is present

Volume of HCl added = 100 cm³ = 0.1 dm³

Concentration of the HCl = 1.20 mol/dm³

Using the formula,

Number of moles = Concentration × Volume

Number of moles of HCl present = 1.20 × 0.1 = 0.120 mole

Since

1 mole of ZnO will completely react with 2 moles of HCl to produce 1 mole of ZnCl₂

Then,

0.06 mole of ZnO will react with the 0.120 mole of HCl to produce 0.06 mole of ZnCl₂

Therefore, maximum number of moles of anhydrous zinc chloride that could be produced is 0.06 mole

Now, for the maximum mass that could be produced

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of ZnCl₂ = 136.286 g/mol

Then,

Mass = 0.06 × 136.286

Mass = 8.17716 g

Mass ≅ 8.18 g

Hence, the maximum mass of anhydrous zinc chloride that could be obtained from the products of the reaction is 8.18 g

Learn more on Stoichiometry here: brainly.com/question/11910892

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What is the molarity of a solution made by dissolving 14.8 g of ammonium hydroxide NH4OH, in enough water to make 250.0 mL of so
kicyunya [14]

Answer:

Molarity= 1.69M

Explanation:

m= 14.8, Mm= 35, V= 0.25dm3, C= ?

Moles = m/M= C×V

Substitute and Simplify

m/M= C×V

14.8/35= C×0.25

C= 1.69M

6 0
3 years ago
Which part of an investigation is only found in an experimental investigation?
pantera1 [17]

Answer:

The correct answer is control group.

Explanation:

A group used in a study or in an experiment, which does not get treatment by the scientists and is used as a foundation to determine the functions of the other tested subjects is known as the control group. The control group is only found in an experimental investigation.  

The group in an experiment, which gets the variable being examined is known as an experimental group. The comparison of an experimental group is done with a control group in order to find the answers in an experiment.  

3 0
3 years ago
Is one mole of zinc the same as one atom of zinc?
Papessa [141]

Answer: No.

Explanation: One mole of zinc is not the same as one atom of zinc. In one mole of zinc, there are approximately 6.022*10^23 atoms of zinc.

3 0
3 years ago
If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show
Talja [164]

Answer:

4.5 g/L.

Explanation:

  • To solve this problem, we must mention Henry's law.
  • Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
  • It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

  • At two different pressures, we have two different solubilities of the gas.

<em>∴ P₁S₂ = P₂S₁.</em>

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

8 0
3 years ago
the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel
shtirl [24]

<u>Answer:</u> The entropy change of the process is 2.0\times 10^2J/K

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{f}}{T}

where,  

\Delta S = Entropy change

n = moles of acetone = 6.3 moles

\Delta H_{f} = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol    (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = -94.7^oC=[273-94.7]=178.3K

Putting values in above equation, we get:

\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K

Hence, the entropy change of the process is 2.0\times 10^2J/K

4 0
4 years ago
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