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harina [27]
3 years ago
8

The average atomic mass of all naturally occurring lithium atoms is 6.941 amu. the two isotopes of lithium are lithium-6 and lit

hium-7. are these isotopes equally common? if not, which is more plentiful in nature, and how do you know?
Chemistry
1 answer:
Gnom [1K]3 years ago
6 0
No, the two isotopes of lithium-6 and lithium-7 are not equally common.  
The more plentiful isotope would be lithium-7. 
This can be easily demonstrated by assuming that both isotopes were equally common. If that were the case, the average atomic mass would be (6 + 7)/2 = 6.5 amu. Now compare that value if they were both equal to the actual value found in nature. The value found in nature is 6.941 amu which is heavier than the 6.5 amu that would happen if they were equally common. Since the natural value is heavier, that means that there has to be more of the heavier isotope than there is of the lighter one. Therefore lithium-7 is more common than lithium-6.
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A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k
Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0831\text{ L bar }mol^{-1}K^{-1}

n = Total number of moles = ?

Putting values in above equation, we get:

5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}         ........(1)

where,

p_A = partial pressure of carbon dioxide = 0.318 bar

p_T = total pressure = 5.57 bar

\chi_A = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571

  • Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, \chi_{N_2}=\frac{0.221}{0.493}=0.448

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0
3 years ago
A bowling ball, a baseball, a tennis ball, and a ping-pong ball are all dropped from a height of 2 meters. Which has the most ki
mr_godi [17]

Answer:

they all have the same amount of kinetic energy

6 0
3 years ago
What is it called when moon is between setting and rising
miv72 [106K]

Answer:

I hope this link helps you.

Explanation:

http://astronomy.swin.edu.au/cosmos/P/Phases

4 0
3 years ago
Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is exce
Iteru [2.4K]

Answer:

10.85 g of water

Explanation:

First we write the balanced chemical equation

3NO_{2} +H_{2}O -->2HNO_{3} +NO

Then we calculate the number of moles of nitric acid produced

n(HNO3) = \frac{mass}{molar mass} =\frac{75.9g}{63.02g/mol}=1.2044 mol

According to the balanced equation, water needed in moles is always half the number of moles of HNO3 produced. So since we will produce 1.2044 mol of HNO3, we will need 0.6022 mol of water. Now to calculate what mass that is:

mass(water)=number of moles*molar mass=0.6022mol*18.02g/mol=10.85g

5 0
3 years ago
If iodine-131 has a half-life of 8 days, how much of a 64.0 g sample of iodine-131 will remain after 32 day?
Mariulka [41]

Answer:

4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.

Explanation:

Half life (t1/2) = 8 days

Original mass (No) = 64 g

Elapsed time (t) = 32 days

Mass remaining (Nt) = ?

Using the half life equation we can obtain the mass remaining (Nt)

Nt = No (1/2) ^t/t1/2

Substituting the values, we have;

Nt = 64 * ( 1/2 ) ^32/8

Nt = 64 * (1/2) ^4

Nt = 64 * 0.0625

Nt = 4 g

So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.

8 0
3 years ago
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