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77julia77 [94]
3 years ago
5

a compound has a percent compostion of carbon equal to 48.8383%, hydrogen equal to 8.1636%, and oxygen equal to 43.1981%. what i

s the mepirical formula
Chemistry
1 answer:
Margarita [4]3 years ago
8 0

Answer:

C₂H₃O

Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 48.8383%

Hydrogen (H) = 8.1636%

Oxygen (O) = 43.1981%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 48.8383%

H = 8.1636%

O = 43.1981%

Divide by their molar mass

C = 48.8383 / 12 = 4.07

H = 8.1636 / 1 = 8.1636

O = 43.1981 / 16 = 2.7

Divide by the smallest

C = 4.07 / 2.7 = 2

H = 8.1636 / 2.7 = 3

O = 2.7 /2.7 = 1

Thus, the empirical formula of the compound is C₂H₃O

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D because O3 means oxygen times 3 times it self so you add another oxygen partial it becomes O4
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2 years ago
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Naily [24]
Following is the balanced <span>radioactive decay series:
</span><span>
Particle/radiations generated during the reaction are labeled in bold at end of reaction. 

Care must be taken that, atomic number and atomic mass number should be balanced in each of these reactions.

1) 92 238U </span>→ <span> 90 234Th + 2 4He(</span>α particle<span>)

A = </span>90 234Th because alpha particle is emitted along with it. So atomic number of daughter element has to be 92 - 2 = 90. This corresponds to Th. <span>

2) 90 234Th  </span>→<span> 91 234Pa + -1 0e (electron)

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4) 92 234U </span>→ 90 230Th + 2 4He (α particle<span>)

</span><span>In this case, 92 234U undergoes nuclear disintegration to generate 90 230Th and alpha particle

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<span>
7) 86 222Rn </span>→<span> 84 218Po + 2 4He </span>(α particle)

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8) 84 218Po </span>→<span> 82 214Pb + 2 4He </span>(α particle)

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<span>
9) 82 214Pb </span>→<span> 83 214Bi + -1 0e (electron)

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I = 
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Answer:

Primer postulado:

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Explanation:

4 0
3 years ago
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dsp73

Answer:

C.

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