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Elena L [17]
3 years ago
10

What is the molecular formula of a component whose molar mass is 88.0 and whose percent composted is 9.1

Chemistry
1 answer:
Shalnov [3]3 years ago
4 0

The molecular formula : C₄H₈O₂

<h3>Further explanation</h3>

Maybe a compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen,so :

mol C :

\tt \dfrac{54.5}{12}=4.54

mol H :

\tt \dfrac{9.1}{1}=9.1

mol O :

\tt \dfrac{36.4}{16}=2.28

Divide by 2.28(the smallest ratio) :

C H : O =

\tt \dfrac{4.54}{2.28}\div \dfrac{9.1}{2.28}\div \dfrac{2.28}{2.28}=2\div 4\div 1

The empirical formula : C₂H₄O

(The empirical formula)n=molecular formula

(2.12+4.1+16)n=88

(44)n=88⇒n=2

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by substitution:

∴ A = 200 mg/ 2^(t/30y)


b) Mass after 90 y :

by  using the previous formula and substitute t by 90 y

A = 200mg/ 2^(90y/30y)

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C) Time for 1 mg remaining:

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3 years ago
Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t
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Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

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I                       0             0

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E                       S             S

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<u>Solubility in 0.0120 M CoBr₂ (S')</u>

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

    CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I                       0           0.0240

C                     +S'           +S'

E                       S'            0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

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