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postnew [5]
3 years ago
14

There are two major isotopes of Lithium. The first has a mass of 6.015 amu and

Chemistry
1 answer:
Marrrta [24]3 years ago
7 0

Answer:

6.9399amu

Explanation:

Isotopes are elements with same atomic number but different atomic masses.

To determine the average atomic mass one must multiply the absolute atomic mass (amu) of each isotope times its absolute atomic mass. This will give the weight average contribution of each isotope. The sum of the weight average contributions gives the final weight average atomic mass of the element. The wt. avg. atomic mass is the value posted on the periodic table.

It is recommended that a table be sketched in the following form ...

Isotope  |    %             fractional amt. X Isotopic mass(amu) => Wt. Avg**.

   Li-6       7.59%    =>      0.0759       x         6.015amu        =>     0.4565

   Li-7      92.41%    =>      0.9241        x         7.016amu        =>     6.4835

*amu => atomic mass units                                 ∑ Li-6 + Li-7  =   6.9399amu

**Wt.Avg. => weight average contribution of isotope = fractional amt x Isotopic mass(amu)

*** Final Wt. Avg. = ∑Wt.Avg. Contributions (value posted of periodic table)

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4 years ago
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Use the chart to determine which type of bond is formed between potassium (K) and chlorine (Cl).
iris [78.8K]
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6 0
4 years ago
Read 2 more answers
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su
anastassius [24]

Answer:

V=27992L=28.00m^3

Explanation:

Hello,

In this case, the combustion of methane is shown below:

CH_4+2O_2\rightarrow CO_2+2H_2O

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3

Best regards.

8 0
4 years ago
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