Answer:
0.189 g.
Explanation:
- This problem is an application on <em>Henry's law.</em>
- Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
- Solubility of the gas ∝ partial pressure
- If we have different solubility at different pressures, we can express Henry's law as:
<em>S₁/P₁ = S₂/P₂,</em>
S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm
S₂ = ??? g/L and P₂ = 5.73 atm
- So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.
<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>
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Answer:
The answer to your question is: C. The specific latent heat of fusion
Explanation:
A. The specific latent heat of vaporization Specific latent heat of vaporization indicates the transition from liquid to vapor, but we are not looking for this definition. This answer is wrong.
B. The specific heat
indicates the amount of heat needed to increase the temperature of water 1°C, so this answer is wrong.
C. The specific latent heat of fusion
. This heat indicate the transition from solid ie to liquid, so this is the right answer.
D. The internal energy measures the energy of the molecules of a substance, so this answer is wrong.
Answer:
92gm
Explanation:
Atomic mass of Mg=24g=1 mole of Mg
∴ 24g =1 mole of Mg contain 6.022×10^23 atom
∴ 6gm contains 246.022×1023×6
=4×6.022×10^23 atoms
Now according to question, there are 6.022×1023 atoms of Na
23gm of Na contains 6.022×10^23 atoms
∴6.022×4×10^23 atoms of Na weighs 23×6.022×10^23×4/6.022×10^23⇒92gm
I am pretty sure that <span>If I were asked to compare matter in solid, liquid, and gaseous states, the statement which would best defined a gas is </span>highest energy, highest molecular motion, and least dense packaging of molecules. I choose this one because it's not sensible to <span>heat CO2 (in case of safety) and in the last option the amount of energy is not satisfying.
Hope it helps!</span>
Answer:
31.5mL
Explanation:
The following were obtained from the question:
C1 (concentration of stock solution) = 2M
V1 (volume of stock solution) =.?
C2 (concentration of diluted solution) = 0.630M
V2 (volume of diluted solution) = 100mL
Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
2 x V1 = 0.630 x 100
Divide both side by 2
V1 = (0.630 x 100) /2
V1 = 31.5mL
Therefore, 31.5mL of 2M solution of FeCl2 required
