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Degger [83]
3 years ago
10

What is the volume of an object with a mass of 31g and density of 444g/ml​

Chemistry
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

<h2>The answer is 0.07 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question

mass = 31 g

density = 444 g/mL

We have

volume =  \frac{31}{444}   \\  = 0.06981...

We have the final answer as

<h3>0.07 mL</h3>

Hope this helps you

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A liquid mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a)
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<u>Answer:</u> The correct answer is Option c.

<u>Explanation:</u>

We are given:

Mass percentage of CH_4 = 20 %

So, mole fraction of CH_4 = 0.2

Mass percentage of C_2H_4 = 30 %

So, mole fraction of C_2H_4 = 0.3

Mass percentage of C_2H_2 = 35 %

So, mole fraction of C_2H_2 = 0.35

Mass percentage of C_2H_2O = 15 %

So, mole fraction of C_2H_2O = 0.15

We know that:

Molar mass of CH_4 = 16 g/mol

Molar mass of C_2H_4 = 28 g/mol

Molar mass of C_2H_2 = 26 g/mol

Molar mass of C_2H_2O = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}

where,

\chi_i = mole fractions of i-th species

m_i = molar masses of i-th species

n_i = number of observations

Putting values in above equation:

\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}

\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75

Hence, the correct answer is Option c.

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