Answer:
Mechanical weathering will cause wearing down of rocks.
Explanation:
All the other choices aren't suitable for a dry climate where sand storms occur.
Answer:
Because of the existence of isotopes.
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:
- Look up the standard reduction potentials for the half reaction·
- Figure out the spontaneous direction.
- 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.
Answer:
There are 4,46x10e-4 moles of CaCl2 (0,000446 mol)
Explanation:
A molar solution indicates the amount of moles of solute (in this case CaCl2) in 1 liter of solution (1000ml).
1000ml -----0,105 mol CaCl2
4, 25 ml----x=(4,25 ml x0,105 mol CaCl2)/1000ml= 4,46x10e-4 mol CaCl2
Answer:
- NH₃ is the limiting reactant.
- Theoretical yield = 120 kg
Explanation:
- 2NH₃ (aq) + CO₂ (aq) → CH₄N₂O (aq) + H₂O (l)
First we <u>convert the given masses of reactants to moles</u>, using their <em>respective molar masses</em>:
- 68.2 kg NH₃ ÷ 17 kg/kmol = 4.01 kmol NH₃
- 105 kg CO₂ ÷ 44 kg/kmol = 2.39 kmol CO₂
2.39 kmol of CO₂ would react completely with (2.39 * 2) 4.78 kmol of NH₃. There are not as many NH₃ kmoles so <u>NH₃ is the limiting reactant.</u>
We <u>calculate how much urea would form with a 100% yield</u>, using the <em>moles of limiting reactant</em>:
- 4.01 kmol NH₃ *
= 2.00 kmol CH₄N₂O
We <u>convert that amount to kg</u>:
- 2.00 kmol CH₄N₂O * 60 kg/kmol = 120 kg CH₄N₂O
Finally we <u>calculate the percent yield</u>:
- 87.5 kg / 120 kg * 100% = 72.9 %