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Ostrovityanka [42]
3 years ago
7

Consider the reaction 2H2S(g)⇌2H2(g)+S2(g),Kp=2.4×10−4 (at 1073 K) A reaction mixture contains 0.111 atm of H2, 0.051 atm of S2,

and 0.566 atm of H2S. Determine how these conditions compare to equilibrium conditions. Match the words in the left column to the appropriate blanks in the sentences on the right.
Chemistry
1 answer:
LenaWriter [7]3 years ago
3 0

Answer:

Q> Kp

The reaction of the system, will be a shift to the left, the side of the reactants.

Explanation:

Step 1: Data given

Kp = 2.4 * 10^-4

Partial pressure H2 =  0.111 atm

Partial pressure S2 =  0.051 atm

Partial pressure H2S = 0.566 atm

Step 2:  The balanced equation

2H2S(g) ⇌ 2H2(g) + S2(g)

Step 3: Calculate Q

Q = (pS2 * (pH2)²) / (pH2S)²

Q = (0.051 * 0.111²) / (0.566²)

Q = 0.00196 =1.96 *10^-3

Q> Kp

Since Q>K, we have more products than reactants (pressure). The reaction of the system, will be a shift to the left, the side of the reactants.

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Step2247 [10]

Answer:

Option D is correct = 8.12 grams of NaCl

Explanation:

Given data:

Moles of sodium chloride = 0.14 mol

Mass of sodium chloride = ?

Solution:

Formula:

Number of moles = mass of NaCl / Molar mass of NaCl

Molar mass of NaCl = 58 g/mol

Now we will put the values in formula.

0.14 mol = Mass of NaCl / 58 g/mol

Mass of NaCl = 0.14 mol  ×  58 g/mol

Mass of NaCl = 8.12 g of NaCl

Thus, 0.14 moles of NaCl contain 8.12 g of NaCl.

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3 years ago
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3 years ago
which of the following requires the most energy to break a bond? a. breaking cl-br bond. b. breaking a n-p bond. c. breaking a o
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6 0
2 years ago
20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc
bezimeni [28]

Answer:

Concentration of the barium ions  = [Ba^{2+}] = 0.4654 M

Concentration of the chloride ions  = [Cl^{-}]=0.9308 M

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride  = 0.1355 M

n=0.1355 M\times  0.04389 L=0.005947 mol

Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

\frac{1}{2}\times 0.05947 mol=0.029735 mol barium hydroxide

Moles of barium hydroxide = 0.029735 mol

Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

=1\times 0.029735 mol= 0.029735 mol

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL  = 63.89 mL = 0.06389 L

Concentration of the barium ions =[Ba^{2+}]

[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M

Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M

8 0
3 years ago
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Answer:

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5 0
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