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ahrayia [7]
3 years ago
12

In a synthesis reaction, one reactant contains 346 J of chemical energy, and one reactant contains 153 J of chemical energy. The

product contains 435 J of chemical energy. Because energy is conserved, what energy change occurs during the reaction?
Chemistry
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

64J of energy must have been released.

Explanation:

Step 1: Data given

One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.

The product contains 435 J of chemical energy.

Step 2:

Since the energy is conserved

Sum of energy of Reactants = Energy of Products

Sum of energy of Reactants = 346 J + 153 J = 499 J

The energy of the product = 435 J

435 < 499

This means energy must have been lost as heat.

Step 3: Calculate heat released

499 J - 435 J = 64 J

64J of energy must have been released.

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What gets reduced in an electrolytic cell made with nickel and copper electrodes?
jekas [21]

Answer:

D. Ni²⁺  

Explanation:

We know at once that the answer cannot be A or C, because Ni and Cu are already in their lowest oxidation states.

The correct answer must be either B or D.

An electrolytic cell is the opposite of a galvanic cell. In the former, the reaction proceeds spontaneously. In the latter, you must force the reaction to occur.  

One strategy to solve this problem is:

  1. Look up the standard reduction potentials for the half reaction·
  2. Figure out the spontaneous direction.
  3. Write the equation in the reverse direction.

1. Standard reduction potentials

                                E°/V

Cu²⁺ + 2e⁻ ⟶ Cu; 0.3419

Ni²⁺ + 2e⁻ ⟶ Ni;  -0.257

2. Galvanic Cell

We reverse the direction of the more negative half cell and add.

                                       <u>E°/V </u>

Ni ⟶ Ni²⁺ + 2e⁻;           0.257

<u>Cu²⁺ + 2e⁻ ⟶ Cu;      </u>   0.3419

Ni + Cu²⁺ ⟶ Cu + Ni²⁺; 0.599

This is the spontaneous direction.

Cu²⁺ is reduced to Cu.

3. Electrochemical cell

                                        <u>E°/V</u>

Ni²⁺ + 2e⁻ ⟶ Ni;           -0.257

<u>Cu ⟶ Cu²⁺ + 2e⁻;        </u> <u>-0.3419</u>

Cu + Ni²⁺ ⟶ Ni + Cu²⁺; -0.599

This is the non-spontaneous direction.

Ni²⁺ is reduced to Ni in the electrolytic cell.

8 0
3 years ago
Calculate the volume of a balloon that can hold 113.4 g of nitrogen dioxide, NO2 gas at STP-
Karolina [17]

Answer:

55.18 L

Explanation:

First we convert 113.4 g of NO₂ into moles, using its molar mass:

  • 113.4 g ÷ 46 g/mol = 2.465 mol

Then we<u> use the PV=nRT formula</u>, where:

  • P = 1atm & T = 273K (This means STP)
  • n = 2.465 mol
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • V = ?

Input the data:

  • 1 atm * V = 2.465 mol * 0.082atm·L·mol⁻¹·K⁻¹ * 273 K

And <u>solve for V</u>:

  • V = 55.18 L
6 0
3 years ago
(06.04 LC)
olga nikolaevna [1]

Answer:

type of particles

that's d

5 0
3 years ago
Read 2 more answers
M → M+ + e- Has M lost or gained an electron?
navik [9.2K]
M has lost an electron.
It's oxidation.
5 0
3 years ago
Read 2 more answers
How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water
Zarrin [17]

Answer:

94.58 g of H_2O

Explanation:

For this question we have to start with the reaction:

H_2~+~O_2~->~H_2O

Now, we can balance the reaction:

2H_2~+~O_2~->~2H_2O

We have the amount of H_2  and the amount of O_2 . Therefore we have to find the limiting reactive, for this, we have to follow a few steps.

1) Find the moles of each reactive, using the molar mass of each compound (H_2~=~2~g/mol~~O_2=~32~g/mol ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for H_2 and "1" for O_2).

<u>Find the moles of each reactive</u>

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}=15.87~mol~H_2

84.0~g~of~O_2\frac{1~mol~of~O_2}{32~g~of~O_2}=2.62~mol~of~O_2

<u>Divide by the coefficient</u>

<u />

\frac{15.87~mol~H_2}{2}=7.94

\frac{2.62~mol~of~O_2}{1}=2.62

The smallest values are for H_2, so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}\frac{2~mol~H_2O}{2~mol~H_2}\frac{18~g~H_2O}{1~mol~H_2O}=94.58~g~H_2O

We have to remember that the molar ratio between H_2O and H_2 is 2:2 and the molar mass of H_2O is 18 g/mol.

6 0
3 years ago
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