Answer:
0.84M
Explanation:
Hello,
At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:
![K_{eq}=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \\K_{eq}=\frac{0.6^2}{0.2*0.2}\\ K_{eq}=9](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNO%5D%5E2_%7Beq%7D%7D%7B%5BN_2%5D_%7Beq%7D%5BO_2%5D_%7Beq%7D%7D%20%5C%5CK_%7Beq%7D%3D%5Cfrac%7B0.6%5E2%7D%7B0.2%2A0.2%7D%5C%5C%20K_%7Beq%7D%3D9)
Now, the new equilibrium condition, taking into account the change x, becomes:
![9=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}}\\9=\frac{[0.9+2x]^2}{[0.2-x][0.2-x]}](https://tex.z-dn.net/?f=9%3D%5Cfrac%7B%5BNO%5D%5E2_%7Beq%7D%7D%7B%5BN_2%5D_%7Beq%7D%5BO_2%5D_%7Beq%7D%7D%5C%5C9%3D%5Cfrac%7B%5B0.9%2B2x%5D%5E2%7D%7B%5B0.2-x%5D%5B0.2-x%5D%7D)
Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:
![\frac{1}{9} =\frac{[N_2]_{eq}[O_2]_{eq}}{[NO]^2_{eq}}\\\frac{1}{9} =\frac{[0.2+x][0.2+x]}{[0.9-2x]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D%20%3D%5Cfrac%7B%5BN_2%5D_%7Beq%7D%5BO_2%5D_%7Beq%7D%7D%7B%5BNO%5D%5E2_%7Beq%7D%7D%5C%5C%5Cfrac%7B1%7D%7B9%7D%20%3D%5Cfrac%7B%5B0.2%2Bx%5D%5B0.2%2Bx%5D%7D%7B%5B0.9-2x%5D%5E2%7D)
Thus, we arrange the equation as:
Finally, the new concentration is:
![[NO]_{eq}=0.9-0.06=0.84M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D0.9-0.06%3D0.84M)
Best regards.
Answer:
Cl^- <S^2-<Sc^3+ <Ca^2+<K^+
Explanation:
We know that ionic radius of ions decreases from right to left in the periodic table. This is because, ionic radii decreases with increase in nuclear charge. This explains why; Sc^3+ <Ca^2+<K^+.
Secondly, even though Cl^- is isoelectronic with S^2-, the size of the nuclear charge in Cl^- is larger compared to that of S^2- . Hence Cl^- is smaller than S^2- in ionic radius owing to increased nuclear attraction in Cl^-.
Answer:
7.82x10^24 molecules of water
Explanation:
H2O=18.015 g/mol Avogadro's #=6.022x10^23 molecules
0.234L x 1000g/1L x 1 mol H2O/18.015 g x 6.022x10^23 = 7.82x10^23 molecules of water
Maybe it’s c I’m not sure
