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boyakko [2]
3 years ago
14

A compound is found to have an empirical formula of CH2O. If its molecular mass is 60.0 g/mol, what is its molecular formula?

Chemistry
1 answer:
Lina20 [59]3 years ago
3 0
Empirical=12.01+2.02+16=30.03
Molecular=60
Molecular/Empirical=2
so, 2(CH2O)
C2H4O2
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The Density equation is:
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Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). Match Term Definition Remo
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Answer:

The answers to the questions are given below.

Explanation:

According to Le Chatelier's principle, if an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.

A. Effective of removing ammonia, NH3.

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2. Effect of removing H2

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Remoing H2 simply means we have more products and less reactant. Therefore, the product will be convert to reactant. Hence, the equilibrium position will shift to the left.

C. Effect of adding a catalyst.

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7 0
3 years ago
A chemist titrates of a hypochlorous acid solution with solution at . Calculate the pH at equivalence. The of hypochlorous acid
const2013 [10]

The question is incomplete, here is the complete question:

A chemist titrates 110.0 mL of a 0.2412 M hypochlorous acid (HCIO) solution with 0.0613 M NaOH solution at 25°C. Calculate the pH at equivalence. The pKa of hypochlorous acid is 7.50. Round your answer to 2 decimal places

<u>Answer:</u> The pH of the solution is 10.09

<u>Explanation:</u>

To calculate the volume of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HClO

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=0.2412M\\V_1=110.0mL\\n_2=1\\M_2=0.0613M\\V_2=?mL

Putting values in above equation, we get:

1\times 0.2412\times 110.0=1\times 0.0613\times V_2\\\\V_2=\frac{1\times 0.2412\times 110.0}{1\times 0.0613}=432.8mL

At equivalence, the number of moles of acid is equal to the number of moles of base. Also, the moles of salt which is NaClO will also be the same.

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

  • <u>For HClO:</u>

Molarity of HClO solution = 0.2412 M

Volume of solution = 110.0 mL

Putting values in equation 1, we get:

0.2412M=\frac{\text{Moles of HClO}\times 1000}{110}\\\\\text{Moles of HClO}=\frac{(0.2412\times 110)}{1000}=0.026532mol

  • <u>For NaClO:</u>

Moles of NaClO = 0.026532 moles

Volume of solution = [432.8 + 110] mL = 542.8 mL

Putting values in above equation, we get:

\text{Molarity of NaClO}=\frac{0.026532\times 1000}{542.8}=0.0489M

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

pH=7+\frac{1}{2}[pK_a+\log C]

where,

pK_a = negative logarithm of weak acid which is hypochlorous acid = 7.50

C = concentration of the salt = 0.0489 M

Putting values in above equation, we get:

pH=7+\frac{1}{2}[7.50+\log (0.0489)]\\\\pH=7+3.09=10.09

Hence, the pH of the solution is 10.09

4 0
3 years ago
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