Full electron configuration of barium: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2
Explanation:

Equilibrium constant of reaction = 
Concentration of NO = ![[NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M](https://tex.z-dn.net/?f=%5BNO%5D%3D%5Cfrac%7B2.69%5Ctimes%2010%5E%7B-2%7D%20mol%7D%7B1%20L%7D%3D2.69%5Ctimes%2010%5E%7B-2%7D%20M)
Concentration of bromine gas = ![[Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D%5Cfrac%7B3.85%5Ctimes%2010%5E%7B-2%7D%20mol%7D%7B1%20L%7D%3D3.85%5Ctimes%2010%5E%7B-2%7D%20M)
Concentration of NOBr gas = ![[Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D%5Cfrac%7B9.56%5Ctimes%2010%5E%7B-2%7D%20mol%7D%7B1%20L%7D%3D9.56%5Ctimes%2010%5E%7B-2%7D%20M)
The reaction quotient is given as:
![Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BNOBr%5D%5E2%7D%7B%5BNO%5D%5E2%5BBr_2%5D%7D%3D%5Cfrac%7B%289.56%5Ctimes%2010%5E%7B-2%7D%20M%29%5E2%7D%7B%282.69%5Ctimes%2010%5E%7B-2%7D%20M%29%5E2%5Ctimes%203.85%5Ctimes%2010%5E%7B-2%7D%20M%7D)


The reaction will go in backward direction in order to achieve an equilibrium state.
1. In order to reach equilibrium NOBr (g) must be produced. False
2. In order to reach equilibrium K must decrease. False
3. In order to reach equilibrium NO must be produced. True
4. Q. is less than K . False
5. The reaction is at equilibrium. No further reaction will occur. False
Answer:
The mass of 2,50 moles of NaCl is 146, 25 g.
Explanation:
First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate the mass of 2.50 moles of compound, making a simple rule of three:
Weight NaCl= Weight Na + Weight Cl= 23 g+ 35,5 g= 58, 5 g/ mol
1 mol ------ 58, 5 g
2,5 mol---x= (2,5 mol x 58, 5 g)/ 1 mol = <u>146, 25 g</u>
I can’t see the picture for some reason