Answer:
3.02e23
Explanation:
carbon molecular weight = 12 g /mol
1 mol = 6.02e23 atoms
your sample, 6 g / 12 g/mol = 0.5 mol
0.5 x 6.02e23 =
We will use this two reaction equation:
H2SO3 + H2O ↔ H3O+ + HSO3- Ka1 = 1.3 x 10^-2
HSO3- + H2O ↔ H3O+ + SO3 2- Ka2= 6.3 x 10^-8
we will use the ICE table for the first equation:
H2SO3 + H2O ↔ H3O+ + HSO3-
initial 0.025 0 0
change -X +X +X
Equ (0.025-X) X X
Ka1 = [H3O+] [HSO3-] / [H2SO3]
1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X
∴ X = 0.0127
when [H3O+] = X
∴[H3O+] = 0.0127 M
and when [HSO3-] = X
∴[HSO3-] = 0.0127 M
and when [H2SO3] = 0.025 - X
∴[H2SO3] = 0.025 - 0.0127
= 0.0123 M
when Kw = [OH-][H3O+]
and Kw = 1.1 x 10^-14 / 0.0127
∴[OH-] = 1.1 x 10^-14 / 0.0127
= 8.66 x 10^-13 M
- by using the ICE table for the second equation:
HSO3- + H2O ↔ H3O+ + SO3 2-
initial 0.0127 0.0127 0
change -X +X +X
Equ (0.0127-X) (0.0127+X) X
when Ka2 = [SO32-] [H3O+] / [HSO3-]
by substitution:
6.3 x 10^-8 = X(0.0127+X) / (0.0127-X)
as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X
6.3 x 10^-8 = 0.0127X /0.0127
∴X = 6.3 x 10^-8
when [SO3 2-] = X
∴[SO32-] = 6.3 x 10^-8
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
Disadvantages of aluminum in building cars include:
On average, aluminum is more expensive than steel, as much as two to three times.
Aluminum is not easily welded. If a steel door or panel suffers a crack, often it cam be welded back together easily, primed and repainted so its difficult to tell any body work has been performed. This cannot be done with aluminum. Steel also can be bent and shaped as needed.
Body shops and even dealer service centers are currently equipped for steel work with technicians skilled in the art of welding and steel fabrication. The cost of switching over to an infra structure designed specifically for all-aluminum vehicles or mass production of aluminum vehicles and components would be high.
