Answer:
a. 82.68 g b. 9.8 min
Explanation:
a. The amount of gold deposited by 5 A current in 2 hrs 15 mins
Since charge Q = It where I = current = 5 A and time = 2 hrs 15 mins = 2 × 60 min + 15 min = 120 + 15 min = 135 min = 135 × 60 s = 8100 s
Q = 5 A × 8100 s
= 40500 C
Also, Q = nF where n = number of moles of gold deposited and F = Faraday's constant = 96500 C
n = Q/F = 40500 C/96500 C = 0.4195 moles ≅ 0.42 mole
Now n = m/M where m = mass of gold and M = molar mass of gold = 197
m = nM
= 0.42 × 197 g
= 82.68 g
b. The time taken for 6g of gold to be deposited.
We first find the number of moles of gold in 6g of gold
Since n = m/M and m = 6 g
n = 6/197 = 0.0305 mole
Q = It = nF
t = nF/I
= 0.0305 mol × 96500 C/5 A
= 2939.09 mol C
= 587.82 s
Changing t to minutes
587.82/60 s = 9.8 min
Answer:
Group VIIIA in which the noble/inert gases are found
Answer:
The mass remains same after and before the reaction
Explanation:
This is because of law of conservation of mass
Which states that
- In a chemical reaction mass is neither created nor destroyed,it remains conserved .
The answer is D. Evaporating water only breaks apart intermolecular bonds like H-bonds and dipole-dipole bonds, and does not change the chemical composition (or intramolecular bonds).
Answer: 128 g/mol
Explanation:
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of the molar mass of its particles.
Mathematically, that is:
Since, you know the ratio of two rates and the molar mass of one gas, you can calculate the molar mass of the other gas.
The molar mass of the oxygen molecule, O₂ = 2×16.0g/mol = 32.0 g/mol.
In the coming equations, I will use 32 g/mol for simplicity of writing.
So, the molecular mass of the unnknown gas is 128 g/mol.
