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:
Question 7 the answer is B
Explanation:
In every nuclear reaction there is always formation of products either by nuclei bombardment or by the elements breaking into two elements of the same mass
Question 8
The answer is A
Nuclear reactions release a huge amount of energy than chemical reactions
Answer:
Atomic number
Explanation:
Because atomic number is unique for each element and never changes
Answer:
- % Cobalt (II) Nitrate = 30.62%
Explanation:
To calculate mass percent, first we need to <u>calculate the total mass of the mixture</u>:
- Mass Water ⇒ 0.350 kg Water = 350 g water
- Mass Ammonia⇒We use ammonia's molar mass⇒5.4 mol * 17 g/mol = 91.8 g
- Mass cobalt (II) nitrate ⇒ 195.0 g
Total Mass = Mass Water + Mass Ammonia + Mass Cobalt Nitrate
- Total Mass = 350 g+ 91.8 g+ 195 g = 636.8 g
To calculate each component's mass percent, we divide its mass by the total mass and multiply by 100:
- % Water ⇒ 350/636.8 * 100% = 54.96%
- % Ammonia ⇒ 91.8/636.8 * 100% = 0.14%
- % Cobalt (II) Nitrate ⇒ 195/636.8 * 100% = 30.62%
Answer:
C: The mobile phase can dissolve the ink and move it up the plate.