The half reaction with the the greater SRP has a greater tendency to gain electrons is the definition of reduction potential when considering a pair of half cell reactions.This reduction potential is measured against hydrogen electrode which is standard electrode.
Answer:
d. 60.8 L
Explanation:
Step 1: Given data
- Heat absorbed (Q): 53.1 J
- External pressure (P): 0.677 atm
- Final volume (V2): 63.2 L
- Change in the internal energy (ΔU): -108.3 J
Step 2: Calculate the work (W) done by the system
We will use the following expression.
ΔU = Q + W
W = ΔU - Q
W = -108.3 J - 53.1 J = -161.4 J
Step 3: Convert W to atm.L
We will use the conversion factor 1 atm.L = 101.325 J.
-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L
Step 4: Calculate the initial volume
First, we will use the following expression.
W = - P × ΔV
ΔV = - W / P
ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L
The initial volume is:
V2 = V1 + ΔV
V1 = V2 - ΔV
V1 = 63.2 L - 2.35 L = 60.8 L
Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
Explanation:
A.
In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.
B.
Equation of the reaction
2NaOH + H2SO4 --> Na2SO4 + 2H2O
Number of moles of H2SO4 = molar concentration × volume
= 0.75 × 0.0105
= 0.007875 moles.
By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875
= 0.01575 moles.
Molar concentration of NaOH = number of moles ÷ volume
= 0.01575 ÷ 0.0175
= 0.9 M of NaOH.
C.) Newton. & it's S.I. Unit of Force.
Hope this helps!