A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0
0 M solution of MSO, and the ight half cell with a 30.0 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 35.0 °C lf right 410 Which electrode will be positive? What voltage will the voltmeter show?
1 answer:
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
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Answer:
The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year
Explanation:
Given that the mass of the carbon 14 at the start = 5 gram
At the end of 5,000 years we will have;
Where
A = The amount of carbon 14 left
A₀ = The starting amount of carbon 14
e = Constant = 2.71828
= The half life
t = The time elapsed = 5000 years
λ = 0.693/ = 0.693/5730 = 0.0001209424
Therefore;
A = 5 × e^(-0.0001209424×5000) = 2.7312 grams
Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left
The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams
The average yearly rate of change of carbon-14 during the first 5000 years is therefore;
2.2688 grams/(5000 years) = 0.0004538 grams per year
The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.