The
chemical reaction is represented as:<span>
2A(g) = B(g) + C(g)
To determine the equilibrium concentration of A, we make use of the equilibrium
constant, Kc, given above. It is expressed as the ratio of the equilibrium
concentrations of the products and the reactants. For this reaction, it is
expressed as:
Kc = [B] [C] / [A]^2
From the problem statement, we are given the following
Kc = 0.035
Volume = 20.0 L
Initial concentrations: [B] = 8.00 mol / 20.0 L = 0.4 M
[C] = 12.00 mol / 20.0 L = 0.6
M
Since the initial reactants are B and C, the reaction is reversed as well as
the Kc.
Kc = [A]^2 / [B][C]
We use the ICE table:
B
C A
I 0.4 0.6
0
C -x -x
+x
------------------------------------------
E 0.4 - x 0.6 - x
x
Kc = x^2 / (0.4-x) (0.6-x) = 0.035
solve for x,
x = 0.07691 = [A]</span>
Answer :
Alpha decay equation :
Beta decay equation :
Gamma decay equation :
Explanation :
- Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.
General representation of alpha decay :
- Beta decay : In this process, a neutron is converted into a proton and an electron. In this decay, the atomic number is increased by 1 unit.
General representation of beta decay :
- Gamma decay : In this process, the unstable nuclei transformed into stable nuclei by releasing gamma rays. In this process, the mass number remains same.
General representation of gamma decay :
Answer:
[NaOH} = 0.4 M
Explanation:
In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.
(H₂SO₄, is considered strong, but the first deprotonation is weak)
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.
In the equivalence point we know mmoles of base = mmoles of acid
Let's finish the excersise with the formula
25 mL . M NaOH = 28.2 mL . 0.355M
M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400
Answer:
<u>4</u>Cr + __302 -->_<u>2</u>Cr203.....