If 6.02×1023 atoms of element Y have a mass of 28.09 g, what is the identity of Y?
1 answer:
Silicon is the element having a mass of 28.09 g
<u>Explanation</u>:
- Silicon is the element having an atomic mass of 28.09 g / mol. So 28.09 g of silicon contains 6.023
10^23 atoms. One mole of each element can produce one mole of compound.
- The Atomic weight of an element can be determined by the number of protons and neutrons present in one atom of that element. So atomic weight expressed in grams always contain the same number of atoms( 6.023
- Avagadro number is the number of atoms of 1 mole of any gas at standard temperature and pressure. It has been determined that 6.023
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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.