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3 years ago

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A certain element consists of two stable isotopes. The first has atomic mass of 7.02 amu and a percent natural abundance of 92.6

%. The second has an atomic mass of 6.02 amu and a percent natural abundance of 7.42%. What is the atomic weight of the element? in amu

1 answer:

statuscvo [17]3 years ago

8 0

<u>Answer:</u> The average atomic mass of the element is 6.95 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

We are given:

Mass of isotope 1 = 7.02 amu

Percentage abundance of isotope 1 = 92.6 %

Fractional abundance of isotope 1 = 0.926

Mass of isotope 2 = 6.02 amu

Percentage abundance of isotope 2 = 7.42 %

Fractional abundance of isotope 2 = 0.0742

Putting values in equation 1, we get:

$\text{Average atomic mass of element}=[(7.02\times 0.926)+(6.02\times 0.0742)]$

$\text{Average atomic mass of element}=6.95amu$

Hence, the average atomic mass of the element is 6.95 amu

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Which statement best describes the relationship between observations and conclusions? A. Observations are based on Conclusions B

irga5000 [103]

Answer:

B. Conclusions are based on observations

Explanation:

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3 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

masha68 [24]

IM IN THAT CLASS TOO

Im abt to do it and ill send u all the answers when im

done, add me on sxxap or instaxxxgram so i can send it to u

my userneamen both: Bellaberrygood

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3 years ago

There are approximately 15 milliliters (mL) in 1 tablespoon (tbsp). Which expression can be used to find the approximate number

Helga [31]

You know 1 tbsp is 15 milliliters so you would multiply 15 and 3 together. 15 x 3 =45
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3 years ago

Read 2 more answers

Both the acetamido (-NHCOCH3) and bromo (-Br) substituents are ortho, para directors. Even though the acetamido group is much la

sweet [91]

Answer:

Explanation has been given below

Explanation:

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A hydrogen bonding can be formed between proton attached with N atom in acetamido group and Cl atom at ortho position. This leads to a formation of a stable 5-membered ring structure.
Hence ortho substitution by Cl is favorable process.
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Structure of hydrogen bonded structure has been shown below.

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3 years ago