Answer:
1. 2 M
2. 2 M
Explanation:
1. Determination of the final concentration.
Initial Volume (V₁) = 2 L
Initial concentration (C₁) = 6 M
Final volume (V₂) = 6 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
6 × 2 = C₂ × 6
12 = C₂ × 6
Divide both side by 6
C₂ = 12 / 6
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
2. Determination of the final concentration.
Initial Volume (V₁) = 0.5 L
Initial concentration (C₁) = 12 M
Final volume (V₂) = 3 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
12 × 0.5 = C₂ × 3
6 = C₂ × 3
Divide both side by 3
C₂ = 6 / 3
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
Given :
Volume of NaCl solution 2.5 L .
Molarity of NaCl solution is 0.070 M .
To Find :
How many moles are present in the solution.
Solution :
Let, n be the number of moles.
We know, molarity is given by :
So,
Therefore, number of moles of NaCl is 0.175 moles.
Answer:
to go against dangerous viral and bacteria
Explanation:
Given the equilibrium reaction: 2 A (aq) + 3 B (aq) <— —> 2 C (aq) + D (aq) and equilibrium concentrations of [A] = 0.60M, [B] = 0.30 M, [C] = 0.10 M and [D] = 0.50 M. The Kc value will be:
a. 1.9 c. 2.4 b. 0.15 d. 0.51
Answer . A
