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Alinara [238K]
3 years ago
10

1. A compound is made up of 11 percent oxygen and 89 percent gold. What would be this compound's

Chemistry
1 answer:
bogdanovich [222]3 years ago
7 0

Answer:

Empirical formula is Au₂O₃

Explanation:

Given data:

Percentage of oxygen = 11%

Percentage of gold = 89%

Empirical formula = ?

Solution:

Solution:

Number of gram atoms of O = 11 / 16 = 0.69

Number of gram atoms of Au = 89 / 197 = 0.45

Atomic ratio:

                Au               :          O    

           0.45/0.45        :      0.69  /0.45

                    1               :           1.5      

Au  : O = 1 : 1.5

Au  : O = 2( 1 : 1.5)

Empirical formula is Au₂O₃.

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The ΔHvap of a certain compound is 49.09 kJ·mol–1 and its ΔSvap is 53.69 J·mol–1·K–1. What is the boiling point of this compound
rusak2 [61]
Hey there !

<span>Convert Joule to KJ :
</span>
1 j ---------------- 0.001 kj
53.69 j ----------- Kj

Kj = 53.69 * 0.001

=>  0.05369 Kj

T = ΔH / <span>ΔS

T = 49.09 / 0.05369

T = 914.32ºC</span>
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5 0
4 years ago
On the reaction below, label the BSA, BSB, CA, and CB. CH3COOH + H2O → CH3COO– + H3O+
Ostrovityanka [42]

Answer:

Acid(BSA) = CH₃COOH

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Explanation:

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Hello,

From my understanding of the question, we are required to identify the

1) Acid

2) Base

3) conjugate acid

4) conjugate base in the reaction

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CB = conjugate base = CH₃COO⁻

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