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puteri [66]
3 years ago
5

A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of

44./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 24.01g water 13.10g Use this information to find the molecular formula of X.
Chemistry
1 answer:
Ne4ueva [31]3 years ago
4 0

Answer:

C3H6.

Explanation:

Data obtained from the question:

Mass of the compound = 8g

Mass of CO2 = 24.01g

Mass of H2O = 13.10g

Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01

=> 12/44 x 24.01 = 6.5g

Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1

=> 2x1/18 x 13.1 = 1.5g

Mass of O in the compound = Mass of compound – (mass of C + Mass of H)

=> 8 – (6.5 + 1.5) = 0

Next, we shall determine the empirical formula of the compound. This is illustrated below:

C = 6.5g

H = 1.

Divide by their molar mass

C = 6.5/12 = 0.54

H = 1.4/1 = 1.

Divide by the smallest

C = 0.54/0.54 = 1

H = 1/0.54 = 2

Therefore, the empirical formula is CH2

Finally, we shall determine the molecular formula as follow:

The molecular formula of a compound is a multiple of the empirical formula.

Molecular formula = [CH2]n

[CH2]n = 44

[12 + (2x1)]n = 44

14n = 44

Divide both side by 14

n = 44/14

n = 3

Molecular formula = [CH2]n = [CH2]3 = C3H6

Therefore, the molecular formula of the compound is C3H6

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0.025M

Explanation:

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Answer:

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Please help me
Wittaler [7]

Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

acid pH can't never be higher than 7.

We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

All the protons come from the bromide and the OH⁻ that come from water.

We can also think [OH⁻] = Kw / [H⁺] so:

[H⁺] = [Br⁻] + Kw / [H⁺]

Now, our unknown is [H⁺]

[H⁺] =  7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]

[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) /  [H⁺]

This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴

(-b +- √(b² - 4ac) / (2a)

[H⁺] = 1.000038751×10⁻⁷

- log [H⁺] = pH → 6.999

A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.

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