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Answer:
Number of moles nitric acid in the cylinder is 400.539g/mol.
Explanation:
From the given,
Weight of empty gas cylinder
= 30.01 g/mol
Number of moles nitric acid =
=?
The mass of nitric acid in the cylinder = 

Number of moles of nitric acid =

Therefore, number of moles nitric acid in the cylinder is 400.539g/mol.
Answer:
B)
Explanation:
It is the theme of the passage.
Answer:
Explanation:
The half-life of K-40 (1.3 billion years) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction
<u>half-lives</u> <u> t/yr </u> <u>Remaining</u>
0 0 1
1 1.3 billion ½
2 2.6 ¼
3 3.9 ⅛
We see that after 2 half-lives, ¼ of the original mass remains.
Conversely, if two half-lives have passed, the original mass must have been four times the mass we have now.
Original mass = 4 × 2.10 g = 
<span>Molar mass(C)= 12.0 g/mol
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol
18g C*1mol C/12 g C = 1.5 mol C
C + O2 → CO2
from reaction 1 mol 1 mol 1 mol
from problem 1.5 mol 1.5 mol 1.5 mol
1.5 mol O2*32 g O2/1 mol O2 = 48 g O2
In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount of O2 that was added for the reaction (probably people who wrote this problem wanted this)
we need (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is </span><span>2.) 54 g.</span>