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adell [148]
3 years ago
9

When copper (Cu) and oxygen (O) combine, two different compounds are formed. The first compound contains 88.8% copper by mass, a

nd the second compound contains 79.9% copper by mass. (Note: Assume a 100-gram sample of each compound.) (a) Show that Cu and O combine with each other in conformity with the law of multiple proportions. The small whole number ratio of the mass of Cu per gram of O in the first compound to the mass of Cu per gram of O in the second compound equals :1. This ratio indicates that Cu and O with each other in conformity with the law of multiple proportions.
Chemistry
1 answer:
marusya05 [52]3 years ago
6 0
1) Chemical reaction 1: 4Cu + O₂ →  2Cu₂O.
n(Cu) = 88,8 ÷ 63,55.
n(Cu) = 1,4.
n(O) = 11,2 ÷ 16.
n(O) = 0,7.
n(Cu) : n(O) = 1,4 : 0,7.
n(Cu) : n(O) = 2 : 1.
Compound is Cu₂O.
2) Chemical reaction 2: 2Cu + O₂ →  2CuO.
n(Cu) = 79,9 ÷ 63,55.
n(Cu) = 1,257.
n(O) = 20,1 ÷ 16.
n(O) = 1,257.
n(Cu) : n(O) = 1,257 : 1,257.
n(Cu) : n(O) = 1 : 1.
Compound is CuO.
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Identify the conjugate acid/base pairs in each of the following equations:
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Answer:

(a) Pair 1: H₂S and HS⁻

    Pair 2: NH₃ and NH₄⁺

(b) Pair 1: HSO₄⁻ and SO₄⁻

    Pair 2: NH₃ and NH₄⁺

(c) Pair 1: HBr and Br⁻

    Pair 2: CH₃O⁻ and CH₃OH

(d)  Pair 1: HNO₃ and NO₃⁻

     Pair 2: H₃O⁺

Explanation:

When an acid loses its proton (H⁺), a conjugate base is produced.

When a base accepts a proton (H⁺), it forms a conjugate acid.

(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.

    NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺

(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.

     The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.

(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.

   CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.

(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.

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6 0
3 years ago
50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.
photoshop1234 [79]

Limiting reactant : O₂

Mass of  N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

\tt =\dfrac{50}{44}=1.136

mol O₂ :

\tt =\dfrac{50}{32}=1.5625

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136    1.5625

1.0416  1.5625    1.0416

0.0944    0          1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

\tt =mol\times MW=1.0416\times 92=95.83~g

7 0
3 years ago
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