According to the first law of thermodynamics, the amount of work done by a heat engine equals the amount of
2 answers:
Answer:
The correct answer is option C.
Explanation:
The first law: thermodynamics:
Q = heat added to the system
w = work done by the system
= Internal energy
The work done by the system is equal to the difference of energy added to the system from the internal energy.This internal energy of the system is the energy which is left unused and given out to the surrounding by the heat engine in the form of thermal energy.
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Answer:- 2.4 mol .
Solution:- It asks to calculate the moles of formed when 1.2 moles of
are reacted.
There is 2:4 mol ratio between and
. So, the moles of reactant are multiplied by the mol ratio to get the moles of
. The calculations are shown below:
= 2.4 mol
So, 2.4 moles of are formed when 1.2 moles of
were reacted.
Answer:
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O
Explanation:
PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O
It will be easiest to balance this equation by the ion-electron method.
1. Write the ionic equation
PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O
2. Eliminate H⁺, H₂O, and spectator ions
PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺
3. Separate the skeleton equation into two half-reactions.
PbO₂ ⟶ Pb²⁺
Mn²⁺ ⟶ MnO₄⁻
4. Balance all atoms other than H and O
Done
5. Balance O by adding water molecules to the deficient side
PbO₂ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻
6. Balance H by adding H⁺ ions to the deficient side.
PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺
7. Balance charge by adding electrons to the deficient side.
PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-
8. Multiply each half-reaction by a number to equalize the electrons transferred.
5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]
2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]
9. Add the two half-reactions.
5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O
<u> 2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻ </u>
5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻
10. Cancel species that occur on each side of the equation
5PbO₂ + 2Mn² + <u>8H₂O</u> + <u>20H⁺</u> + <u>10e⁻</u> ⟶ 5Pb²⁺ + 2MnO₄⁻ + <u>10H₂O</u> + <u>16H⁺</u> + <u>10e⁻ </u>
becomes
5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O
11. Add the missing spectator ions
5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O
+ 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻ +2SO₄²⁻ + 6NO₃⁻ + 2H⁺
becomes
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O
12. Check that all atoms are balanced.
Everything checks. The balanced equation is
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O