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3 years ago

You start with 12.94 g sample of a mixture containing SiO2, NaCl, and NH4Cl. If after subliming

Chemistry

1 answer:

Tema [17]3 years ago

8 0

<h3>Answer:</h3>

83.23 %

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of the sample mixture is 12.94 g
Mass of the remaining mixture after sublimation is 2.17 g

We are required to get the mass percent of NH₄Cl in the original sample

<h3>Step 1 : Calculate mass of NH₄Cl in the original sample;</h3>

Mass of NH₄Cl = Total mass of the mixture - remaining mass after sublimation

= 12.94 g - 2.17 g

= 10.77 g

<h3>Step 2: Percent by mass of NH₄Cl in the original sample</h3>

% of NH₄Cl by mass =(Mass of NH₄Cl÷Total mass of original sample)×100

= (10.77 g ÷ 12.94 g) × 100

= 83.23 %

Therefore, the percent by mass of NH₄Cl in the original sample is 83.23%

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The balanced redox reactions for the sequential reduction of vanadium are given below.

Minchanka [31]

Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

$2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)$ $2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)$

$2V^{3+} (aq)+ Zn (s)\rightarrow 2V^{2+}(aq)+Zn^{2+}(aq)$

On adding above equations, we get the following combined equation.

$2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)$

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

$12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}$

From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.

Let us use this as a conversion factor to find the moles of Zn.

$0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn$

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

$0.00059895mol Zn\times \frac{65.38gZn}{1molZn}=0.0392gZn$

We need 0.0392 grams of Zn metal to completely reduce vanadium.

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3 years ago

A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2

sukhopar [10]

Answer:

a) $\triangle G^{0} = 7.31 kJ/mol$

b) $K_{-1} = 0.0594 m^{-1} s^{-1}$

Explanation:

Equation of reaction:

$2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)$

Initial pressure 3 1 0

Pressure change 2P 1P 2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

$K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }$

$K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}$

Standard free energy:

$\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol$

b) value of k−1 at 27 °C, i.e. 300K

$K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}$

$K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}$

$K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2} }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}$

6 0

3 years ago