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3 years ago

7

For the following reaction, find the value of Q and predict the direction of change, given that a 1L flask initially contains 2

moles S8, 2 moles SF6, and 2 moles F2.
1/8 S8 (s) + 3 F2 (g) ⇄ SF6 (g) Kc = 0.425

(A) Q = K, system is at equilibrium
(B) Q < K, reaction will make more reactants
(C) Q < K, reaction will make more products
(D) Q > K, reaction will make more reactants
(E) Q > K, reaction will make more products

1 answer:

Tresset [83]3 years ago

5 0

Answer:

C) Q < K, reaction will make more products

Explanation:

1/8 S8(s) + 3 F2(g) ↔ SF6(g)

∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³

∴ Q = [ SF6 ] / [ F2 ]³

∴ [ SF6 ] = 2 mol/L

∴ [ F2 ] = 2 mol/L

⇒ Q = ( 2 ) / ( 2³)

⇒ Q = 0.25

⇒ Q < K, reaction will make more products

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0.042 moles of Hydrogen evolved

<h3>Further explanation</h3>

Given

I = 1.5 A

t = 1.5 hr = 5400 s

Required

Number of Hydrogen evolved

Solution

Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.

Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)

Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)

So at the cathode H₂ gas is produced

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$\tt mol~e^-=\dfrac{Q}{96500}$

Q = i.t

Q = 1.5 x 5400

Q = 8100 C

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3 years ago

Sodium oxide reacts with water to make sodium hydroxide.

Vlada [557]

Answer:

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Hydrocarbons do not dissolve in concentrated sulfuric acid, but methyl benzoate does. Explain this difference and write an equat

finlep [7]

Answer:

See explanation

Explanation:

For a substance to dissolve in another, there must be some sort of interaction between the substances.

Recall that like dissolves like. That is, polar substances dissolve polar substances and non polar substances dissolve nonpolar substances.

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3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
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Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

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Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

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Darya [45]

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