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alexandr1967 [171]
3 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
Chemistry
1 answer:
vovikov84 [41]3 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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Consider the following reaction: NO( g) + SO 3( g) ⇌ NO 2( g) + SO 2( g) A reaction mixture initially contains 0.86 atm NO and 0
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Answer:

The equilibrium pressure of NO2 is 0.084 atm

Explanation:

Step 1: Data given

A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3.

Kp = 0.0118

Step 2: The balanced equation

NO( g) + SO3( g) ⇌ NO2( g) + SO2( g)

Step 3: The initial pressures

p(NO) = 0.86 atm

p(SO3) = 0.86 atm

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p(SO2) = 0 atm

Step 4: The pressure at the equilibrium

For 1 mol NO we need 1 mol SO3 to produce 1 mol NO2 and 1 mol SO2

p(NO) = 0.86 -x atm

p(SO3) = 0.86 -xatm

p(NO2) = x atm

p(SO2) = x atm

Step 5: Define Kp

Kp = ((pNO2)*(pSO2)) / ((pNO)*(pSO3))

Kp = 0.0118 = x²/(0.86 - x)²

X = 0.08427

p(NO) = 0.86 -0.08427 = 0.77573 atm

p(SO3) = 0.86 -0.08427 = 0.77573 atm

p(NO2) = 0.08427 atm

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The equilibrium pressure of NO2 is 0.08427 atm ≈ 0.084 atm

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What is the percent mass oxygen in calcium carbonate (CaCo3)?
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  <span>Step 1 is to determine the mass of each part 
Mass of Ca is 40.08 g 
Mass of C is 12.01 g 
Mass of O is 16.00 x 3 = 48.00 g 
Step 2 is to determine the total mass of the compound 
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g 

Step 3 is to determine the % of each part using the following formula: 
Mass of part / total mass x 100 = 

40.08 / 100.09 x 100 = 40.04 % Ca 

12.01 / 100.09 x 100 = 12.00 % C 

48.00 / 100.09 x 100 = 47.96 % O 

Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :) 
40.04 
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