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Darina [25.2K]
3 years ago
13

An unknown compound contains only carbon and hydrogen. Combustion analysis of the compound produces 0.2845 g of carbon dioxide a

nd 0.1451 g of water. What is the empirical formula of the compound? Remember: the carbon from the compound is used to form carbon dioxide and the hydrogen from the compound is used to form water. Also remember there are two hydrogens for each water.
Chemistry
1 answer:
Len [333]3 years ago
5 0

Answer:

C2H5

Explanation:

As the question says, the carbon from the compound is used to form carbon dioxide and the hydrogen is used to form the water, if the analysis of the compound produces 0.2845 g of and 0.1451 g and we know the molar mass of CO2 and H2O(44 and 18 (g/mol), respectively), it is possible to calculate the number of moles of CO2 and H2O.

<u>Number of moles= g/molar mass </u>

Number of moles of CO2= 0.2845g/ (44 g/mol)= 0.0065 moles of CO2, if one mole of CO2 has 1 mole of C, so <em>there are 0.0065 moles of C  </em>

Number of moles of H2O= 0.1451/(18 g/mol)= 0.0081 moles of H2O. There are two hydrogens for each water, so there are 0.0162.

<em>The empirical formula shows the ratio of the elements in the compound</em>, so it is possible to know the ratio between C and H dividing each number of moles by the smallest numer

C= 0.0065/0.0065=1

H=0.0162/0.0065=2.5

Then if we multiply by two, the empirical formula is C2H5, it is because it can´t be represented by non integer numbers.

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Explanation:

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V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}

<em>B) Radius</em>

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V = \frac{ 4}{3 }\pi r^{3}

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3 0
3 years ago
Help needed ASAP, I will mark your answer as brainliest.
evablogger [386]

Answer:

b

Explanation:

b

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