Answer:
84.8 mL
Explanation:
From the question given above, the following data were obtained:
Mass of CuNO₃ = 3.53 g
Molarity of CuNO₃ = 0.330 M
Volume of solution =?
Next, we shall determine the number of mole in 3.53 g of CuNO₃. This can be obtained as follow:
Mass of CuNO₃ = 3.53 g
Molar mass of CuNO₃ = 63.5 + 14 + (16×3)
= 63.5 + 14 + 48
= 125.5 g/mol
Mole of CuNO₃ =?
Mole = mass / Molar mass
Mole of CuNO₃ = 3.53 / 125.5
Mole of CuNO₃ = 0.028 moles
Next, we shall determine the volume of the solution. This can be obtained as follow:
Molarity of CuNO₃ = 0.330 M
Mole of CuNO₃ = 0.028 moles
Volume of solution =?
Molarity = mole /Volume
0.330 = 0.028 / Volume
Cross multiply
0.330 × Volume = 0.028
Divide both side by 0.330
Volume = 0.028 / 0.330
Volume = 0.0848 L
Finally, we shall convert 0.0848 L to millilitres (mL). This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.0848 L = 0.0848 L × 1000 mL / 1 L
0.0848 L = 84.8 mL
Therefore, the volume of the solution is 84.8 mL.
The cost of gasoline for the trip would cost $57.28 because you first divide 320 by 20 which gives you 16 than after that you do 16 multiplied by 3.58 which gives you $57.28 as your final product
Answer:
0.26 moles
Explanation:
Number of moles = mass/molar mass
Mass of SiO2 = 15.5g
Molar mass of SiO2 = 60.08g/mol
Number of moles = 15.5g/60.08g/mol
=0.257
= 0.26 moles
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.
