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alexandr1967 [171]

2 years ago

10

Element X has two isotopes. One has a mass of 45.0 amu and is 45.0% abundant. The other has a mass of 47.0 amu and is 55.0% abun

dant. What is the average atomic mass of element X?*
0 46.1 amu
O 36.5 amu
0 43.8 amu
0 48.9 amu

1 answer:

vovikov84 [41]2 years ago

7 0

The <em>average</em><em> </em>atomic mass of element X that has two isotopes (one of 45.0 amu mass and 45% abundance and the other of 47.0 amu mass and 55.0% abundance) is 46.1 amu (option 1).

The <em>average </em>atomic mass (A) of element X can be calculated as follows:

$A = m_{1}\%_{1} + m_{1}\%_{1}$ (1)

Where:

m₁: is the mass of isotope 1 = 45.0 amu

m₂: is the <em>mass </em>of <em>isotope </em>2 = 47.0 amu

%₁: is the abundance percent of <em>isotope </em>1 = 45.0 %

%₂: is the <em>abundance percent</em> of <em>isotope </em>1 = 55.0 %

Hence, the <u>average atomic mass</u> is (eq 1):

$A = m_{1}\%_{1} + m_{1}\%_{1} = 45.0 amu*45.0\% + 47.0 amu*55.0\% = 45.0 amu*0.45 + 47.0 amu*0.55 = 46.1 amu$

Therefore, the <em>average </em>atomic mass of element X is 46.1 amu (option 1).

Find more about atomic mass here:

brainly.com/question/3187640?referrer=searchResults

brainly.com/question/17338557?referrer=searchResults

I hope it helps you!

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A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a

Semenov [28]

<u>Answer:</u> The correct answer is Option d.

<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

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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 d.

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3 years ago

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Answer:

Explanation:

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= .25 moles

mole of CO₂

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= .1818 moles

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$\frac{8}{64}$

= .125 moles

Total moles of gas

= .5568 moles.

total volume of gas mixture

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3 years ago

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Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

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where,

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Putting values in above equation, we get:

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