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juin [17]
3 years ago
13

Calculate the average atomic mass of argon to two decimal places, given the following relative atomic masses and abundances of e

ach of the isotopes: argon-36 (35.97 u; 0.337%), argon-38 (37.96 u; 0.063%), and argon-40 (39.96 u; 99.600%)
Chemistry
1 answer:
user100 [1]3 years ago
4 0

Answer:

39.95 u

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})+(\frac {\%\ of\ the\ third\ isotope}{100}\times {Mass\ of\ the\ third\ isotope})

Given that:

<u>For first isotope, Argon-36: </u>

% = 0.337 %

Mass = 35.97 u

<u>For second isotope, Argon-38: </u>

% = 0.063 %

Mass = 37.96 u

<u>For third isotope, Argon-40: </u>

% = 99.600 %

Mass = 39.96 u

Thus,  

Average\ atomic\ mass=\frac{0.337}{100}\times {35.97}+\frac{0.063}{100}\times {37.96}+\frac{99.600}{100}\times {39.96}=0.1212189+0.0239148+39.80016=39.9452937

<u>Average atomic mass = 39.95 u</u>

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Explanation:

The given data is as follows.

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      height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)

Density = 13.534 g/mL = 13.534 g/mL \times \frac{10^{6}cm^{3}}{1 m^{3}} \times \frac{1 kg}{1000 g}

                = 13534 kg/m^{3}

The relation between pressure and atmospheric pressure is as follows.

             P = P_{atm} + \rho gh

Putting the given values into the above formula as follows.

            P = P_{atm} + \rho gh

               = 98700 Pa + 13534 \times 9.81 \times 0.03 m

               = 102683.05 Pa

               = 102.68 kPa

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3 years ago
Which shows the conversion of 8.93 × 10-2 grams to milligrams?
Allushta [10]
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1. Why were atoms given the name of "atom" which means "unable to be cut?"
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3 years ago
What is the atomic number of arsenic (As)?<br> O A. 33<br> O B. 15<br> C. 75<br> D. 4<br> SUB
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Answer:

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Explanation:

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How many moles of nitrogen gas are there in 16.8 L of this gas at STP?
sweet-ann [11.9K]
According to Avogadro's Law, same volume of any gas at standard temperature and pressure will occupy same volume. And one mole of any Ideal gas occupies 22.4 dm³ (1 dm³ = 1 L).

Data Given:

                  n = moles = ?
                  V = Volume = 16.8 L

Solution:
               
As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....

                           = ( 1 mole × 16.8 L) ÷ 22.4 L
                    
                           = 0.75 moles

Result:
           
16.8 L of Nitrogen gas will contain 0.75 moles at standard temperature and pressure.
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