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2 years ago

A. What is the pH of a solution with a [H+] of 6.8 x 10^-11?

Chemistry

1 answer:

inysia [295]2 years ago

6 0

Answer:

[H+] = 6.8×10^-11

so, pH = - log[H+]

= - log [6.8×10^-11]

= -(-10.167)

= 10.167

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Given the equation: HCl + Na2SO4 → NaCl + H2SO4, if you start with 8 moles of hydrochloric acid, how many grams of sulfuric acid

Simora [160]

Answer:

392g sulfuric acid are produced

Explanation:

Based on the balanced equation:

2HCl + Na2SO4 → 2NaCl + H2SO4

2 moles of HCl produce 1 mole of sulfuric acid

To solve the problem we need to find the moles of sulfuric acid produced based on the chemical equation. Then, using its molar mass -Molar mass H2SO4 = 98g/mol- we can find the mass of sulfuric acid produced:

Moles sulfuric acid:

8mol HCl * (1mol H2SO4 / 2mol HCl) = 4 mol H2SO4

Mass sulfuric acid:

4mol H2SO4 * (98g / mol) =

392g sulfuric acid are produced

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3 years ago

For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car

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Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Na₂CO₃: 1 mole
Ca(NO₃)₂: 1 mole
CaCO₃: 1 mole
NaNO₃: 2 mole

Being the molar mass of the compounds:

Na₂CO₃: 106 g/mole
Ca(NO₃)₂: 164 g/mole
CaCO₃: 100 g/mole
NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

Na₂CO₃: 1 mole* 106 g/mole= 106 g
Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
CaCO₃: 1 mole* 100 g/mole= 100 g
NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

$mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}$

mass of CaCO₃= 74.81 grams

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

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