1) What is the mole ratio of D to A in the generic chemical reaction? (2 points) Page 390 helps with understanding mole ratios.
1 answer:
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Answer:
29Si has a natural abundance of 4.68%.
30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.
Explanation:
The atomic mass of silicon is given by:
Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃
Where:
Si: atomic mass of silicon (28.086)
Si²⁸: relative atomic mass of 28Si (27.97693)
A₁: natural abundance of 28Si (92.23%)
Si²⁹: relative atomic mass of 29Si (28.97649)
A₂: natural abundance of 29Si
Si³⁰: relative atomic mass of 30Si
A₃: natural abundance of 30Si
We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.
We have to set up a system of three equations in three unknowns:
Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃
A₃=0.6592×A₂
A₁+A₂+A₃=1
First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:
A₁+A₂+0.6592×A₂=1
A₁+1.6592×A₂=1
1.6592×A₂=1-A₁
A₂==0.0468
Then, we find the value of A₃:
A₃=0.6592×A₂
A₃=0.6592×0.0468=0.0309
Finally, we find the value of Si³⁰ in the first equation:
Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃
28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309
28.086=27.15922+Si³⁰×0.0309
28.086-27.15922=Si³⁰×0.0309
=Si³⁰
Si³⁰=29.99288
Answer:
0.011 mol/L
Explanation:
This can be solved with something called an ICE table.
I = initial
C = change
E = equilibrium
Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.
x amount of N₂ reacts. Since the stoichiometry is 1:1, x amount of O₂ also reacts. This produces 2x of NO.
After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.
Here it is in table form:
Now we can use the equilibrium constant:
Kc = [NO]² / ( [N₂] [O₂] )
Substituting:
0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )
Solving:
0.10 = (2x)² / (0.04 - x)²
√0.10 = 2x / (0.04 - x)
(√0.10) (0.04 - x) = 2x
(√0.10)(0.04) - (√0.10)x = 2x
(√0.10)(0.04) = 2x + (√0.10)x
(√0.10)(0.04) = (2 + √0.10)x
x = (√0.10)(0.04) / (2 + √0.10)
x = 0.0055
At equilibrium, the concentration of NO is 2x. So the answer is:
[NO] = 2x
[NO] = 0.011
The equilibrium concentration of NO is 0.011 mol/L.