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3 years ago

65.5g of copper reacts with __g of oxygen to produce 81g copper(I) oxide

Chemistry

1 answer:

Nikitich [7]3 years ago

7 0

Answer:

grams O₂(g) ≅ 15.9 g O₂(g) (3 sig. figs.)

Explanation:

2Cu + O₂ => 2CuO

Convert given to moles, solve by equation rxn ratio then convert to grams.

Moles CuO formed = 81.0 g / 81.5 g·mol⁻¹ = 0.994 mol CuO

0.994 mol CuO requires 0.994 mol Cu = 0.994 mol Cu x 64 g·mol⁻¹ = 63.6 g Cu => from this, the amount of O₂ needed = 1/2(0.994 mol) O₂(g) = 0.497 mol O₂(g) = (0.497 mol O₂(g))(32 g·mol⁻¹) = 15.904 g O₂(g) ≅ 15.9 g O₂(g) (3 sig. figs.)

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2 years ago

A scientist prepared an aqueous solution of a 0.45 M weak acid. The pH of the solution was 2.72. What is the percentage ionizati

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Answer:

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2.72 = - log[H⁺]

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