Answer:
(i) specific heat
(ii) latent heat of vaporization
(iii) latent heat of fusion
Explanation:
i. Q = mcΔT; identify c.
Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.
ii. Q = mLvapor; identify Lvapor
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.
iii. Q = mLfusion; identify Lfusion
Here, Q is the heat, m is the mass and L is the latent heat of fusion.
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.
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Answer:</h3>
Chlorine gas (Cl₂)
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Explanation:</h3>
- According to the Graham's law of diffusion, the diffusion rate of a gas is inversely proportional to the square root of its density or molar mass.
- Therefore, a lighter gas will diffuse faster at a given temperature compared to a heavy gas.
- Consequently, the heavier a gas is then the denser it is and the slower it diffuses at a given temperature and vice versa.
In this case we are given gases, CI₂
, H₂,He and Ne.
- We are required to identify the gas that will diffuse at the slowest rate.
- In other words we are required to determine the heaviest gas.
Looking at the molar mass of the gases given;
Cl₂- 70.91 g/mol
H₂- 2.02 g/mol
He - 4.00 g/mol
Ne- 20.18 g/mol
Therefore, chlorine gas is the heaviest and thus will diffuse at the slowest rate among the choices given.
Answer:
it’s mass was greater than when it started
Explanation:
When a metal is coated with another metal, the plating metal deposits on the plated metal. Usually, the plating metal functions as the anode while the plated metal functions on the cathode. The anode metal is oxidized and reduced at the cathode and become deposited on the cathode material. This increases the mass of the cathode. Hence the mass of the silver/gold product is greater than the mass of silver at the beginning of the electroplating process.