Answer: 406 hours
Explanation:

where Q= quantity of electricity in coloumbs
I = current in amperes = 39.5 A
t= time in seconds = ?
The deposition of copper at cathode is represented by:

Coloumb of electricity deposits 1 mole of copper
i.e. 63.5 g of copper is deposited by = 193000 Coloumb
Thus 19.0 kg or 19000 g of copper is deposited by =
Coloumb

(1hour=3600s)
Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A
Answer:

Explanation:
Mole ratio


Moles of CO₂

Use formula to find mass

Relative molecular mass of CO₂ = 12 + 16 × 2 = 44


B. Heating up the reaction will increase the entropy of a reaction.
<h3>
What is entropy?</h3>
Entropy is the measure of the degree of disorderliness of a system.
Entropy is also the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.
S = ΔH/T
where;
- S is entropy
- ΔH is energy input
- T is temperature
Entropy increases in reactions in which the total number of product molecules is greater than the total number of reactant molecules.
However, entropy increases as temperature increases. Thus, heating up the reaction will increase the entropy of a reaction.
Learn more about entropy here: brainly.com/question/6364271
#SPJ1
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234





So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left

Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.
Answer:
22.73s
Explanation:
The reaction is a second order reaction, we know this by observing the unit of the slope.
rate constant = k = 0.056 M-1s-1
the initial concentration of BrO- [A]o = 0.80 M
time = ?
Final concentration [A]t= one-half of 0.80 M = 0.40M
1 / [A]t = kt + 1 / [A]o
1 / 0.40 = 0.056 * t + 1 / 0.80
t = (2.5 - 1.25) / 0.056
t = 22.73s