Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
<span>pm stands for picometer and picometers are units which can be used to measure really tiny distances. One picometer is equal to 10^{-12} meters. We know that one centimeter is equal to 10^{-2} m so there are 10^2 cm per meter.
We can change the distance d = 115 pm to units of centimeters.
d = (115 pm) x (10^{-12}m / pm) x (10^2 cm / m)
d = 115 x 10^{-10} cm = 1.15 x 10^{-8} cm
The distance in centimeters is 1.15 x 10^{-8} cm</span>
Assume 1 liter = 1 kilogram of water = 1000 grams of water.
Part A)
MW of hydrogen is 1.008g/mol, and oxygen is 16.00g/mol.
Find the MW of water by
2*(1.008) + (16.00) = 18.016g/mol.
Convert 1000g H2O to moles :
(1000g H2O)*(1mol H2O / 18.016g H2O) = 55.51 mol
Part B)
Using the answer from part A and Avogadro's number:
(55.51mol)*(6.022*10^23) =
3.343*10^25 molecules.
Hope this is helpful
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Answer:
41.63g
Explanation:
Given parameters:
Volume of CaCl₂ = 500mL = 0.5L
Concentration = 0.75mol/L
Unknown:
Mass of the solute needed = ?
Solution:
The mass of the solute can be derived using the expression below;
Mass = number of moles x molar mass
But,
Number of moles = Concentration x Volume
So;
Mass = Concentration x Volume x molar mas
Molar mass of CaCl₂ = 40 + 2(35.5) = 111g/mol
Mass = 0.75 x 0.5 x 111 = 41.63g
