3 years ago

Scribing the Phases of the Moon

Chemistry

2 answers:

barxatty [35]3 years ago

5 0

Answer:

1, 3, 6

Explanation:

algol133 years ago

5 0

Answer:

the first, second, third, and fifth are true.

Explanation:

Hope this helps! (and its okay if you cant find the science subject i struggled to find it too when i first started.)

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A 100g of sample of a compound is combusted in excess oxygen and the products are 2.492g of CO2 and 0.6495 of H2O. Determine the

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The empirical formula : C₁₁O₁₄O₃

<h3>Further explanation</h3>

The assumption of the compound consists of C, H, and O

mass of C in CO₂ =

$\tt \dfrac{12}{44}\times 2.492=0.680~g$

mass of H in H₂O =

$\tt \dfrac{2.1}{18}\times 0.6495=0.072~g$

mass of O :

mass sample-(mass C + mass H)

$\tt 1-(0.68+0.072)=0.248`g$

mol of C :

$\tt \dfrac{0.68}{12}=0.056$

mol of H :

$\tt \dfrac{0.072}{1}=0.072$

mol of O :

$\tt \dfrac{0.248}{16}=0.0155$

divide by 0.0155(the lowest ratio)

C : H : O ⇒

$\tt \dfrac{0.056}{0.0155}\div \dfrac{0.072}{0.0155}\div \dfrac{0.0155}{0.0155}=3.6\div 4.6\div 1\\\\\dfrac{11}{3}\div \dfrac{14}{3}\div \dfrac{3}{3}=11:14:3$

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