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Maslowich
2 years ago
10

Calculation: If a compound is 14.40% hydrogen by mass and 85.60% carbon by mass, what is the empirical formula?

Chemistry
1 answer:
IgorLugansk [536]2 years ago
3 0

Answer:

CH2

Explanation:

We are given that a compound is 14.40% hydrogen and 85.60% carbon.

Let the mass of the substance be 100g.

Mass of the hydrogen: 14.40% of 100 = 14.40 g

Mass of the carbon: 85.60% of 100 = 85.60 g

Now, let's find the moles of hydrogen:

14.40 g*\frac{1 mole}{1gH} = 14.40 mole

Moles of carbon:

85.60g*\frac{1 mol}{12gC} = 7.13mol

Let's put these in a ratio and simplify:

7.13 mole C: 14.40 mole H

1 mole C: 2 mole H

Therefore, the empirical formula of this compound is CH2.

Hope this helps!! If you have any questions about my work, please let me know in the comments!

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Answer:

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Explanation:

From the question we are told that:

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Y=B⇌C,ΔG= -29.7 kJ/mol

Z=C⇌D,ΔG= 8.10 kJ/mol

Since

Hess Law

The law states that the total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken.

Therefore

Generally the equation for the Reaction is mathematically given by

T = +1 * X +1 * Y +1 *Z

Therefore the free energy, ΔG is

\triangle G=1 * \triangle G*X +1 * \triangle G*Y +1 * \triangle G *Z

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\triangle G= -6.7 KJ/mol

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A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
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Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

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Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

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Step 4: Calculate the limiting reactant

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