Answer:
![[NO]=\frac{k_{-1}}{k_1} [N_2O_2]](https://tex.z-dn.net/?f=%5BNO%5D%3D%5Cfrac%7Bk_%7B-1%7D%7D%7Bk_1%7D%20%5BN_2O_2%5D)
Explanation:
Hello!
In this case, since the reaction may be assumed in chemical equilibrium, we can write up the rate law as shown below:
![r=-k_1[NO]+k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=r%3D-k_1%5BNO%5D%2Bk_%7B-1%7D%5BN_2O_2%5D)
However, since the rate of reaction at equilibrium is zero, due to the fact that the concentrations remains the same, we can write:
![0=-k_1[NO]+k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=0%3D-k_1%5BNO%5D%2Bk_%7B-1%7D%5BN_2O_2%5D)
Which can be also written as:
![k_1[NO]=k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=k_1%5BNO%5D%3Dk_%7B-1%7D%5BN_2O_2%5D)
Then, we solve for the concentration of NO to obtain:
Best regards!
Answer:
See below
Explanation:
Hello there!
Electronegativity is the atom's tendency to attract electrons in a chemical bond
There are two trends to electronegativity:
- Electronegativity increases from bottom to top in a group (Li has a greater electronegativity than Fr, for example)
- Electronegativity increases from left to right across a period (the further right the group, the greater the electronegativity)
Looking at a period table, Sb, Sn, Te, and I are all in the same period, so we'll need to decide which element has the highest electronegativity based on the group.
- Sn belongs to group 4A (group 14)
- Sb belongs to group 5A (group 15)
- Te belongs to group 6A (group 16)
- I belongs to group 7A (group 17)
As I belongs to group 7A, the group that is the farthest right based off of the options given, I has the highest electronegativity
Hope this helps!
Answer:
personally I like the mortal kombat 11 better
Answer:
63.9%
Explanation:
moral mass of x (# x ions) -divided by- Moral mass of compound.
% of cl = (35.5g) (2 cl ions) -divided by- 111.1g × 100
%= 63.9%
