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WINSTONCH [101]
3 years ago
15

Classify each of the following as a strong acid or a weak acid and indicate how each should be written in aqueous solution. Clas

sify ... In solution this acid should be written as: weak 1. hydrocyanic acid H3O CN- _______ 2. hydrobromic acid
Chemistry
1 answer:
JulijaS [17]3 years ago
7 0

Answer:

HCN, weak acid

H⁺, Br⁻, strong acid

Explanation:

Hydrocyanic acid is a weak acid, according to the following equation.

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Thus, it should be written in the undissociated form (HCN).

Hydrobromic acid is a strong acid, according to the following equation.

HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)

Thus, it should be written in the ionic form (H⁺, Br⁻).

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A) An ionic compound dissolved in a polar solvent

Explanation:

Potassium Chloride's chemical formula is KCl. Most ionic compounds are formed between a nonmetal and a metal. In this case, potassium is acting as the metal and chloride is the nonmetal. Water is a polar solvent due to the electronegativity of the oxygen in the molecule creating a partial negative pole, leaving the hydrogen atoms partially positive. Hence, A is your best answer.

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In a laboratory experiment the reaction of 3.0 mol of H2 with 2.0 mol of I2 produced 1.0 mole of HI. Determine theoratical yield
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Answer:

Theoretical yield of HI is 512 g.

The percent yield for this reaction is 25%.

Explanation:

H_2+I_2\rightarrow 2HI

Moles of hydrogen gas = 3.0 moles

Moles of iodine gas  = 2.0 moles

According to reaction 1 mol of hydrogen gas reacts with 1 mol of iodine gas.

Then 3.0 moles of hydrogen gas reacts with 3.0 mol of iodine gas. But there are 2.0 moles of iodine gas. Hence,Iodine is a limiting reagent. The production of HI will depend upon iodine gas moles.

According to reaction , 1 mol of iodine gas gives 2 moles of HI.

Then 2 moles of iodine gas will give:

\frac{2}{1}\times 2 mol=2 mol of HI

Theoretically we will get 4 moles of HI.

Theoretical yield of HI =  4 mol × 128 g/mol= 512 g

Experimental yield of HI = 1.0 mol

= 1 mol × 128 g/mol= 128 g

\%yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}} \times 100

\%yield=\frac{128 g}{512 g}\times 100=25\%

The percent yield for this reaction is 25%.

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