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Tanya [424]
3 years ago
9

Calculate the hydronium ion concentration in an aqueous solution with a ph of 9.85 at 25°c.

Chemistry
2 answers:
UkoKoshka [18]3 years ago
8 0

Answer:

1.41 × 10⁻¹⁰ M

Explanation:

We have a solution with a pH of 9.85 at 25 °C. We can calculate the concentration of H⁺ using the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH

[H⁺] = antilog -9.85

[H⁺] = 1.41 × 10⁻¹⁰ M

H⁺ is usually associated with water molecules forming hydronium ions.

H⁺ + H₂O → H₃O⁺

Then, the concentration of H₃O⁺ ions is 1.41 × 10⁻¹⁰ M.

nordsb [41]3 years ago
8 0

Answer:

The hydronium ion concentration for a solution with pH 9.85 = 1.41 *10^-10M

Explanation:

Step 1: data given

pH = 9.85

Temperature = 25.0 °C

the hydronium ion concentration = [H3O+] = [H+]

Step 2: Calculate the hydronium ion concentration

[H3O+] = [H+]

pH = -log[H+]

9.85 = -log[H+]

[H+] = 10^-9.85

[H+] = 1.41 *10^-10M = [H3O+]

The hydronium ion concentration for a solution with pH 9.85 = 1.41 *10^-10M

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zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

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A chemist measures the amount of hydrogen gas produced during an experiment. She finds that 264. g of hydrogen gas is produced.
Lesechka [4]

Answer:

The answer is 130.953 g of hydrogen gas.

Explanation:

Hydrogen gas is formed by two atoms of hydrogen (H), so its molecular formula is H₂. We can calculate is molecular weight as the product of the molar mass of H (1.008 g/mol):

Molecular weight H₂= molar mass of H x 2= 1.008 g/mol x 2= 2.01568 g

Finally, we obtain the number of mol of H₂ there is in the produced gas mass (264 g) by using the molecular weight as follows:

mass= 264 g x 1 mol H₂/2.01568 g= 130.9731703 g

The final mass rounded to 3 significant digits is 130.973 g

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