a) when Kc = concentration of products / concentration of reactants
So according to the reaction equation:
Br2(g) + Cl2(g) → 2BrCl(g)
∴ Kc =[BrCl] ^2 / [Br2][Cl2]
b) when q = [BrCl]^2 / [Br2][Cl2]
and we have [BrCl] = 3 m
[Br2] = 1 m
[Cl2] = 1 m
So by substitution:
q= 3^2 / 1*1 = 9
- and we can see that q > Kc
the reaction is not at equilibrium that means there are more products and the reaction shifts to the left to increase the reactants and decrease the products to achieve equilibrium.
C) by using ICE table:
Br2(g) + Cl2(g) → 2BrCl (g)
initial 1 1 3
change -X -X +X
Equ (1-X) (1-X) (3+X)
when Kc = [Brcl]^2/[Cl2][Br2]
by substitution:
7 = (3+X)^2 / (1+X) (1+X) by solving this equation for X
∴X = 0.215
so at equilibrium:
∴ [Br2] = [Cl2] = 1-0.215 = 0.785 m
[BrCl] = 3+0.215 = 3.215 m
The gram formula mass is Molar mass. The mass of 1.0 moles is :
3) 48.0 g
Answer:

Explanation:
This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.


The units of milliliters (mL) cancel.



The temperature changes to <u>232.9 degrees Celsius.</u>
<span>The pH of a vinegar solution is 4.15. To find the H+ concentration of the solution use the following equation -log(H+)=pH.
Insert the pH into the equation to get, -log(H+) = 4.15
Rearrange the equation to get, 10^(-4.15) = H+
Finally, you can solve for H+.
The hydrogen ion concentration of the vinegar solution is .0000708 M.</span>