Answer:
is the solubility of nitrogen gas in a diver's blood.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
where,
= Henry's constant = 
= partial pressure of nitrogen
(Raoult's law)
is the solubility of nitrogen gas in a diver's blood.
Answer: 17) d. 
18. c. The empirical formula of a compound can be twice the molecular formula.
Explanation:
Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.
Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.
To calculate the molecular formula, we need to find the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
The empirical mass can be calculated from empirical formula and molar mass must be known.
17. Thus the empirical formula of should be 
18. The molecular formula will either be same as empirical formula or is a whole number multiple of empirical formula. Thus the empirical formula of a compound can never be twice the molecular formula.
I would agree with the second one, not the first. You can't always see the chemical reaction, and it isn't always sudden. But the second claim is true.
Answer:
Hey, I will try my best
so as you know volcanos can be very destructive and cause mass death and deforestation. but nature needs volcanos because they are part of a cycle. The lava that comes from volcanos can feed various plants which will start to grow after an eruption. these plants will process the lava rock into food which feeds other plants. and soon enough you have a whole forest after all the lava rock has been processed.
spiky bob your answerer
Answer:
The energies of combustion (per gram) for hydrogen and methane are as follows: Methane = 82.5 kJ/g; Hydrogen = 162 kJ/g
<em>Note: The question is incomplete. The complete question is given below:</em>
To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/℃. When a 1.00-g sample of methane gas burned with
<em>excess oxygen in the calorimeter, the temperature increased by 7.3℃. When a 1.00 g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3°C. Compare the energies of combustion (per gram) for hydrogen and methane.</em>
Explanation:
From the equation of the first law of thermodynamics, ΔU = Q + W
Since there is no expansion work in the bomb calorimeter, ΔU = Q
But Q = CΔT
where C is heat capacity of the bomb calorimeter = 11.3
kJ/ºC; ΔT = temperature change
For combustion of methane gas:
Q per gram = (
11.3
kJ/ºC * 7.3°C)/1.0g
Q = 83 kJ/g
For combustion of hydrogen gas:
Q per gram = (
11.3
kJ/ºC * 14.3°C)/1.0g
Q = 162 kJ/g
