Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
Answer:
1219.5 kj/mol
Explanation:
To reach this result, you must use the formula:
ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)
ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).
The BE values are:
BE C = C: 839 kj / mol
BE C-H: 413 Kj / mol
BE O = O: 495 kj / mol
BE C = O = 799 Kj / mol
BE O-H = 463 kj / mol
Now you must replace the values in the above equation, the result of which will be:
ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol
Q is unlike K value it describes the reaction that is not at equilibrium.
by considering this reaction:
aA+ bB⇄ cC
and our reaction is:
Br2 + Cl2 ⇄ 2 BrCl
According to Q low:
Q= concentration of products/concentration of reactants
but this equation in the gaseous or aqueous states only.
∴ Q = [BrCl]^2 / [Br2] [Cl2]
and we have [Br2] = 0.00366 m [Cl2]= 0.000672 m [BrCl] = 0.00415 m
by substitution:
= [0.00415]^2 / ( [0.00366] * [0.000672])
∴ Q = 7
Answer:
Mass of CuSO4.H2O obtained: 
Explanation:
The molecular weight of the salt is: 
<u>In the solution</u>: 12.5 g of CuSO4
In moles: 
<u>Mass of wate</u>r:
5 moles of water per mol of salt: 
Mass of CuSO4.H2O obtained: 
A) 5.2 x 10^2
B) 86.
C) 6.4 x 10^3
D) 5.0
E) 22.
F) 0.89
