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julsineya [31]
3 years ago
8

Question 8 4 pts What would be the resulting molarity of a solution made by dissolving 31.3 grams of Ca(OH)2 in enough water to

make a 1050-milliliter solution? Show all of the work needed to solve this problem.
Chemistry
2 answers:
Tcecarenko [31]3 years ago
4 0

Answer : The molarity of the solution is, 0.4028 mole/L

Explanation : Given,

Mass of Ca(OH)_2 = 31.3 g

Molar mass of Ca(OH)_2 = 74 g/mole

Volume of solution = 1050 ml

Molarity : It is defined as the moles of solute present in one liter of solution.

Formula used :

Molarity=\frac{\text{Mass of }Ca(OH)_2\times 1000}{\text{Molar mass of }Ca(OH)_2\times \text{volume of solution in ml}}

Now put all the given values in this formula, we get:

Molarity=\frac{31.3g\times 1000}{74g/mole\times 1050ml}=0.4028mole/L

Therefore, the molarity of the solution is, 0.4028 mole/L

Gennadij [26K]3 years ago
3 0
Answer is: <span>molarity of a solution is 0,401 M.
</span>m(Ca(OH)₂) = 31,3 g.
n(Ca(OH)₂) = m(Ca(OH)₂) ÷ M(Ca(OH)₂).
n(Ca(OH)₂) = 31,3 g ÷ 74 g/mol.
n(Ca(OH)₂) = 0,422 mol.
V(solution) = 1050 mL · 0,001 L/mL = 1,050 L.
c(Ca(OH)₂) = n(Ca(OH)₂) ÷ V(solution).
c(Ca(OH)₂) = 0,422 mol ÷ 1,050 L.
n(Ca(OH)₂) = 0,401 mol/L = 0,401 M.
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Answer:

The correct answer is

B.

\Delta H=2(1072)+498 -4(799)

Explanation:

Enthalpy of reaction :

It is the amount of energy released/absorbed when one mole of the substance is formed from the reactant at a constant pressure.

The enthalpy of a reaction can be calculated using :

\Delta H=\Delta H_{reactants}-\Delta H_{products}

2CO+O_{2}\rightarrow 2CO_{2}

\Delta H_{reactants}=2(C\equiv O)+O=O

\Delta H_{reactants}=2(1072)+498

H_{products}=2(2\times C=O)

H_{products}=4\times 799

\Delta H=\Delta H_{reactants}-\Delta H_{products}

\Delta H=2(1072)+498 -4(799)

units = \frac{kJ}{mole}

Please note that :

The carbon monoxide , CO should be taken as C triple bond O. Not C=O .

So , the bond energy =1072 is used

\Delta H=2(1072)+498 -4(799)

8 0
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Contact [7]

Answer:

Half-reactions:

Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻

Net ionic equation:

2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺

Explanation:

The Cr³⁺ is reduced to Cr²⁺:

<h3>Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>

Zn is oxidized to Zn²⁺:

<h3>Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>

Twice the reduction of Cr:

2Cr³⁺ + 2e⁻ → 2Cr²⁺

Now this reaction + Oxidation of Zn:

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<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>

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alekssr [168]
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ch sulfide (ZnS) occurs in the zinc blende crystal structure, (a) If 254 g of ZnS contains 170 g of Zn, what is the mass ratio o
Aleonysh [2.5K]

Answer:

The mass ratio of zinc to sulfide is 85:42.

2.5559 kg of Zn are in 3.82 kg of ZnS.

Explanation:

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Explanation:

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