Answer:
1.05 V
Explanation:
Since;
E°cell= E°cathode- E°anode
E°cathode= -0.40 V
E°anode= -1.45 V
E°cell= -0.40-(-1.45) = 1.05 V
Equation of the process;
2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)
n= 8 electrons transferred
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]
Since log 1=0
Ecell= E°cell= 1.05 V
We need to first add both of the solution volumes together 35+115=150. Now we can divide the volume of the ethanol by the total volume 35/150=.233. To double check we can multiply the total volume by the percentage of ethanol by volume we got as a solution 150x.233=35. So the percentage by volume of ethanol in the solution is .233x100=23.3%.
Explanation:
Mole ratio of Oxygen to Hydrogen gas = 1 : 2.
If we use 3.0 moles of oxygen gas, we would need 3.0 * 2 = 6.0 mol of hydrogen gas.
However we only have 4.2 mol of hydrogen. Therefore hydrogen is limiting and oxygen is in excess. (B)
Answer: hydrogen is the limiting reactant.
Explanation:
We have the equation
.
This means that for every mole of nitrogen consumed, 3 moles of hydrogen are consumed.
- Considering the nitrogen, the reaction can occur 0.50 times.
- Considering the hydrogen, the reaction can occur 1.8/3 = 0.6 times.
Therefore, <u>hydrogen</u> is the limiting reactant.
Answer:
2600 kg has 4 significant figures
Explanation:
Any different number to zero is a significant figure
For example, 1.8 → 2 significant figures
When you have 0 in the middle of two numbers, we consider it as a significant number
For example, 3.02 → 3 significant figures
When you have 0 on the left in the measure, we do not consider as a sgnificant figures
For example 0.0010 → 2 significant figures (10)
3.0 → 2 significant figures (0 is on the right, not the left)