Answer:
you aready did it ? would love to help
Explanation:
Answer:
0.07 M
Explanation:
Given data:
Volume of H₂SO₄ = 35.00 mL (35/1000 = 0.035 L)
Molarity of KOH =0.1222 M
Volume of KOH = 42.22 mL (42.22/1000 = 0.0422 L)
Molarity /concentration of H₂SO₄ = ?
Solution:
Chemical equation:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
Number of moles of KOH:
Molarity = number of moles / volume in L
0.1222 M = number of moles / 0.0422 L
0.0422 L×0.1222 M = number of moles
Number of moles = 0.005 mol
Now we will compare the moles of KOH with H₂SO₄ from balance chemical equation.
KOH : H₂SO₄
2 : 1
0.005 : 1/2×0.005 = 0.0025 mol
Concentration of H₂SO₄:
Molarity = number of moles / volume in L
Molarity = 0.0025 mol / 0.035 L
Molarity = 0.07 M
Answer:
Mole fraction O₂ = 452 mmHg/ 780mmHg → 0.579
Explanation:
Mole fraction is defined as the quotient between Partial Pressure of a gas, in a mixture and the Total Pressure from the mixture
Partial pressure of a gas/ Total pressure
Mole fraction O₂ = 452 mmHg/ 780mmHg → 0.579
Be careful, because mole fraction doesn't have units
Answer:
The entropy change in the environment is 3.62x10²⁶.
Explanation:
The entropy change can be calculated using the following equation:
Where:
Q: is the energy transferred = 5.0 MJ
: is the Boltzmann constant = 1.38x10⁻²³ J/K
: is the initial temperature = 1000 K
: is the final temperature = 500 K
Hence, the entropy change is:
Therefore, the entropy change in the environment is 3.62x10²⁶.
I hope it helps you!
Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O
<h3>Further explanation</h3>
Given
MnO4- + Br- = MnO2 + BrO3-
Required
The half-reaction
Solution
In acidic conditions :
1. Add the coefficient
2. Equalization O atoms (add H₂O) on the O-deficient side.
3. Equalization H atoms (add H⁺ ) on the H -deficient side. .
4. Equalization of charge (add electrons (e) )
5. Equalizing the number of electrons and then adding the two half -reactions together
Oxidation : Br- → BrO3-
Reduction : MnO4- → MnO2
Br- + 3H2O → BrO3-
MnO4- → MnO2 + 2H2O
Br- + 3H2O → BrO3- + 6H+
MnO4- + 4H+ → MnO2 + 2H2O
Br- + 3H2O → BrO3- + 6H+ + 6e- x1
MnO4- + 4H+ + 3e- → MnO2 + 2H2O x2
- Equalizing the number of electrons and then adding the two half -reactions together
Br- + 3H2O → BrO3- + 6H+ + 6e-
2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O
Br- + 2MnO4- + 3H2O + 8H+ + 6e- → BrO3- + 2MnO2 + 6H+ + 4H2O + 6e-
Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O