How muchmuch hydrogen gas evolvedwhen 1.5 current is passed through water for 1.5 hours?
1 answer:
0.042 moles of Hydrogen evolved
<h3>Further explanation</h3>Given
I = 1.5 A
t = 1.5 hr = 5400 s
Required
Number of Hydrogen evolved
Solution
Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.
Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)
Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻
Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)
So at the cathode H₂ gas is produced
Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C
Q = i.t
Q = 1.5 x 5400
Q = 8100 C
mol e⁻ = 8100 : 96500 = 0.084
From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042
You might be interested in
<u>Answer:</u> The final temperature of the coffee is 43.9°C
<u>Explanation:</u>
To calculate the final temperature, we use the equation:
where,
q = heat released =
m = mass of water = 10.0 grams
C = specific heat capacity of water = 4.184 J/g°C
= final temperature = ?
= initial temperature = 20°C
Putting values in above equation, we get:
Hence, the final temperature of the coffee is 43.9°C
The question is incomplete. The complete question is:
The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
570 years
Explanation:
The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.
