Answer:
Max. work done in 60 g of copper plated out is 200472.14 J
Explanation:
Given cell reaction is:
Standard reduction potential of Zn electrode (
) is 0.763 V.
Standard reduction potential of Cu electrode (
) is -0.337 V.
Copper acts as cathode and Zinc acts as anode.
Cell potential (E) = E° cathode - E° anode
= 0.763 - (-0.337)
= 1.10 V
formula for the work done is as follows:
Here, n is no. of electron involved in the reaction.
F(Faraday's constant) = 96500
In the given reaction, n = 2
Therefore, 212300 J work is done by reducting 1 mol of copper.
Copper given is 60 g.
Molecular mass of copper is 63.54 g/mol.
Max. work done in 60 g of copper plated out is:
The biotic factors are bacteria soil, dead leaves, and stream water.
<h3>What is an abiotic factor?</h3>
An abiotic factor is a non-living part of an ecosystem that shapes its environment.
Biotic and abiotic factors make up a community via interaction.
Biotic factors are considered living things (having "life") while abiotic factors are simply non-living things.
The dead leaves of plants are an abiotic factor, the bacteria in soil are living matter and stream flowing, etc.
Hence, the biotic factors are bacteria soil, dead leaves, and stream water.
Learn more about the abiotic factor here:
brainly.com/question/12689972
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Answer:
This solution is quite lengthy
Total system = nRT
n was solved to be 0.02575
nH20 = 0.2x0.02575
= 0.00515
Nair = 0.0206
PH20 = 0.19999
Pair = 1-0.19999
= 0.80001
At 15⁰c
Pair = 0.4786atm
I used antoine's equation to get pressure
The pressure = 0.50
2. Moles of water vapor = 0.0007084
Moles of condensed water = 0.0044416
Grams of condensed water = 0.07994
Please refer to attachment. All solution is in there.
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
