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PIT_PIT [208]
1 year ago
12

Compare molecular & covalent network crystal solid?​

Chemistry
1 answer:
Daniel [21]1 year ago
8 0

Answer:

Molecular solids and covalent network solids are two types of solid compounds. The key difference between molecular solid and covalent network solid is that <em>molecular solid forms due to the action of Van der Waal forces </em>where as <em>covalent network solid forms due to the action of covalent chemical bonds.</em>

hope this helps

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Make a prediction about the relative boiling points of the Noble gases
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5 0
2 years ago
A compound contains only carbon, hydrogen, and oxygen. combustion of 11.75 mg of the compound yields 17.61 mg co2 and 4.81 mg h2
Ymorist [56]

Number of moles is defined as the ratio of given  mass in g to the molar mass.

First, convert the given mass of carbon dioxide in mg to g:

1 mg = 0.001 g

17.61 mg = 0.01761 g

Number of moles of carbon dioxide = \frac{0.01761 g}{44.01 g/mol}

= 0.0004001 mol

Mass of carbon  = number of moles of carbon dioxide \times molar mass of carbon

= 0.0004001 mol\times 12.011 g/mol

= 0.004806 g

Number of moles of water= \frac{0.00481 g}{18 g/mol}

= 2.672\times 10^{-4}

Since, water contains two hydrogen atoms. Thus,

Moles of hydrogen = 2\times 2.672\times 10^{-4}

= 5.34\times 10^{-4}

Mass of hydrogen = 5.34\times 10^{-4}\times \times 1.008 g/mol

= 5.34\times 10^{-4} g

Mass of oxygen = 0.001175-(5.38\times 10^{-4}g+0.004806 g)

= 0.006405 g

Number of moles of oxygen = \frac{0.006405 g}{15.999 g/mol}

= 0.000400

Now,

C_{0.0004001}  H_{0.000534}  O_{0.000400}

Divide the smallest number to get the whole number,

C_{\frac{0.0004001}{0.000400}}  H_{\frac{0.000534}{0.000400}}  O_{\frac{0.000400}{0.000400}}

we get,

C_{1}  H_{1.33}  O_{1}

Now, multiply all the subscript by 3 to get the whole number,

C_{3}     H_{4}      O_{3}   (empirical fomula)

Molar mass of the compound  =3\times 12.011 g/mol+4\times 1.008 g/mol+3\times 15.999 g/mol

= 88.062 g/mol

Divide given molar mass of the compound with the molar mass of the compound.

=\frac{176.1 g/mol}{88.062 g/mol}

= 1.999\simeq 2

Thus, multiply the subscripts of empirical formula by 2 to get the molecular formula, we get:

C_{6}H_{8}O_{6}

Hence, empirical formula is C_{3}H_{4}O_{3} and molecular formula is C_{6}H_{8}O_{6}



8 0
3 years ago
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HCI, so C HCI is an acid
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