What is an acid?

If 832J of energy is required to raise the temperature of a sample of aluminum from 20.0° C to 97.0° C, what mass is the sample

LiRa [457]

Data:
Q (Amount of heat) = 832 J
m (mass) = ?
c (Specific heat) = 0.90 J/(g × ° C)
T (final) = 97 ºC
To (initial) = 20 ºC
ΔT = T - To → ΔT = 97 - 20 → ΔT = 77 ºC

Formula:
Q = m*c*ΔT

Solving:
Q = m*c*ΔT
832 = m*0.90*77
832 = 69.3m
69.3m = 832
$m = \frac{832}{69.3}$
$\boxed{\boxed{m \approx 12.00\:g}}\end{array}}\qquad\quad\checkmark$

5 0

3 years ago

Which best describes the transition from gas to liquid?

Nina [5.8K]

A - its condensation and gas particles have a higher kinetic energy

8 0

3 years ago

Read 2 more answers

Please help me equalize: PbO2 + MnSO4 + HNO3 = HMnO4 + PbSO4 + Pb(NO3)2 + H2O

Feliz [49]

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

2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

10. Cancel species that occur on each side of the equation

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

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.

$\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{Pb} & 5 & 5\\\text{O} & 36 & 36\\\text{S} & 2 & 2\\\text{H} & 6 & 6\\\text{N} & 6 & 6\\\end{array}$

Everything checks. The balanced equation is

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

7 0

4 years ago

Some1 please answer these questions in stuck

attashe74 [19]

the reactions at the cathode:

1. the reduced active metal is water, other than that the metal will be reduced

2. H⁺ of the acid will be reduced

the reactions at the anode:

1. if the electrodes are not inert then the metal is oxidized

2. If inert then:

a. OH⁻ from the base will be oxidized

b. The halogen metal will oxidize

1. HCl

H⁺ of the acid will be reduced

Cathode: 2H⁺(aq) + 2e⁻ ⇒ H₂(g)

The halogen metal will oxidize

Anode: 2Cl⁻(aq) ⇒ Cl₂(g) + 2e⁻

2. NaNO₃

Na : active metal, water reduced

Cathode: 2H₂O + 2e ⇒ 2OH- + H₂

NO₃: oxyacid, water will be oxidized

Anode: 2H₂O ⇒ 4 H + + O₂ + 4e

3. CuCl

Cu : not active metal

Cathode: Cu²⁺ (aq) + 2e ⇒ Cu (s)

Cl : The halogen metal will oxidize

Anode: 2Cl⁻ (aq) ⇒ Cl₂ (g) + 2e

4 0

3 years ago

Hellpppppoppopppppppp

Kipish [7]

Answer:

C

Explanation:

it's always colder at night and close to night especially near water and that's when wind tends to blow the most

8 0

4 years ago

Read 2 more answers