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diamong [38]
3 years ago
14

What is the pH of a solution with [H+] = 1.25 mc015-1.jpg 10–10 M? Use mc015-2.jpg. –10.1 –9.90 7.90 9.90

Chemistry
2 answers:
kipiarov [429]3 years ago
6 0
We use the relationship:
pH = -log[H⁺]
pH = -log(1.25 x 10⁻¹⁰)

pH = 9.9
Mademuasel [1]3 years ago
4 0

Answer : Option D) 9.90

Explanation : The pH of the solution can be determined by the formula as

pH = - log [H^{+}]

Here, the concentration of [H^{+}] = 1.25 X 10^{-10} M

So, substituting this value in above formula;

we get pH = -log  [ 1.25 X 10^{-10} M]

Therefore, pH = 9.90.

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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
3 years ago
In an experiment to determine how to make sulfur trioxide, a chemist combines 32.0 g of sulfur with 50.0 g of oxygen. She finds
LuckyWell [14K]

Answer:

The chemist needs to react 40 g of sulfur with 60 g of oxygen to make 100 g of sulfur trioxide.

Explanation:

2S (s) + 3O₂ (g) → 2SO₃ (g)

64g    + 96g     →  160 g

32g    + 48g     →   80 g

   x     +     y      →  100 g

   

1 mol SO₃ ___ 80g

     n _______ 100g

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1 mol S ___ 32 g

1,25 mol S __ 40 g

1 mol O₂ ___ 32 g

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4 0
3 years ago
A compound is 57.1% oxygen and the remainder carbon. using oxygen as the standard (16 amu) and assuming a 1:1 atom ratio, calcul
Delicious77 [7]
Let's start with the amount given in percent. Let our basis be 100 grams of compound. So, that means that in this amount, 57.1 g is oxygen and 100-57.1=42.9 g is carbon. Since there is 1:1 atom ratio, it also means that moles oxygen = moles carbon.

Moles = Mass/Relative Mass
Let x be the relative mass of oxygen

57.1/16 = 42.9/x
Solving for x,
<em>x = 12.02 amu</em>
4 0
3 years ago
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