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
Answer: 45.983 g CBr₄
Explanation:
To convert from moles to grams, you know that we will need molar mass and Avogadro's number.
Avogadro's number: 6.022×10²³ molecules/mol
Molar mass: 331.627 grams/mol
Now that we have what we need, you can use these to solve for grams. 
Our final answer is 45.983 g CBr₄.
Answer:
0.0055 mol of N2O5 will remay after 7 min.
Explanation:
The reaction follows a first-order.
Let the concentration of N2O5 after 7 min be y
Rate = Ky = change in concentration of N2O5/time
K is rate constant = 6.82×10^-3 s^-1
Initial concentration of N2O5 = number of moles/volume = 2.1×10^-2/1.8 = 0.0117 M
Change in concentration = 0.0117 - y
Time = 7 min = 7×60 = 420 s
6.82×10^-3y = 0.0117 - y/420
0.0117 - y = 420×6.82×10^-3y
0.0117 - y = 2.8644y
0.0117 = 2.8644y + y
0.0117 = 3.8644y
y = 0.0117/3.8644 = 0.00303 M
Number of moles of N2O5 left = y × volume = 0.00303 × 1.8 = 0.0055 mol (to 2 significant digits)
Oxygen has six valence (outer-shell) electrons so therefore gains two more electrons to form the O-2 ion Its electron configuration is: 1s2 2s2 2p6 or Ne
If this helped mark Brainliest!
That depends. there are 2 possible answers.
H
C - C = C - H gives a different answer on the right than on the left.
One the left side, the second Carbon is attached to a double bond and has but one hydrogen attached to it.
The Carbon on the right of the double bond has 2
H
C- C = C - H
H
I'm not sure what you should put. It's one of those things that I would repeat my argument and submit it.
