When E° cell is an electrochemical cell which comprises of two half cells.
So,
when we have the balanced equation of this half cell :
Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V
and we have also this balanced equation of this half cell :
Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V
so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)
when E° = E°2 - E°1
∴E° =0.8 - (-1.66)
= 2.46 V
∴ the correct answer is 2.46 V
Answer:
It takes 86 days take to cover half of the lake
Explanation:
In the day #1, the amount of the algae is X,
In the day #2 is 2X
In the day #3 is 2*2*X = X*2²
...
In the day #n the amount of the algae is X*2^(n-1)
Assuming X = 1m³. In the day 87, the area infected was:
1m³*2^(87-1)
7.74x10²⁵m³ is the total area of the lake
the half of this amount is 3.87x10²⁵m³
The time transcurred is:
3.87x10²⁵m³ = 1m³*2^(n-1)
Multiplying for 5 in each side:
ln (3.87x10²⁵) = ln (2^(n-1))
58.9175 = n-1 * 0.6931
85 = n-1
86 = n
<h3>It takes 86 days take to cover half of the lake</h3>
Answer:
90%
Explanation:
Percentage yield = ?
Theoretical yield = 50g
Actual yield = 45g
To calculate the percentage yield of a compound, we'll have to use the formula of percentage yield which is the ratio between the actual yield to theoretical multiplied by 100
Percentage yield = (actual yield / theoretical yield) × 100
Percentage yield = (45 / 50) × 100
Percentage yield = 0.9 × 100
Percentage yield = 90%
The percentage yield of the substance is 90%
Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
Solution : Given,
Density of solution = 
Molar mass of sulfuric acid (solute) = 98.079 g/mole
98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.
Mass of sulfuric acid (solute) = 98.0 g
Mass of solution = 100 g
Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g
First we have to calculate the volume of solution.
Now we have to calculate the molarity of solution.
Now we have to calculate the molality of the solution.
Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
