The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2
so 1mole of NH3 will give 1 mole of NO2
43.9 grams of NH3 contains 2.58 moles
so 2.58 moles will be produced of NO2
which is 118.7 grams this the amount of oxygen that is used.
Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
mass = 0.508 g, Volume = 0.175 L
Temperature = (25 + 273) K = 298 K, P = 1 atm
As per the ideal gas law, PV = nRT.
where, n = no. of moles = 
Hence, putting all the given values into the ideal gas equation as follows.
PV = 
1 atm \times 0.175 L = 
= 71.02 g
As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of 
is 70.8 g/mol or 71 g/mol (approx).
Thus, we can conclude that the gas is most likely chlorine.
The pressure exerted when both gases are put together in a single 1 liter container is 5 atm.
<h3>What is pressure?</h3>
Pressure is the force exerted by any object on another object.
Given that, a and b separate 1 liter containers and exert pressure of 2 atm and 3 atm respectively.
When both gases a and b exert together, the pressure then
2 atm + 3 atm = 5 atm.
Thus, the pressure exerted when both gases are put together in a single 1 liter container is 5 atm.
Learn more about pressure
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The moles of oxygen gas (O2) that is needed is 4 moles
Explanation
2H2 +O2 → 2H2O
The moles of O2 is determined using the mole ratio of H2:O2
that is from equation above H2:O2 is 2:1
If the moles of H2 is 8 moles therefore the moles of O2
= 8 moles x 1/2 = 4 moles
As per the given chemical formula- Na2CO3.10H2O, one mole of the chemical compound contains 13 moles of oxygen atoms. Hence
Number of moles of oxygen atoms in one mole of Na2CO3.10H2O = 13
number of moles of oxygen atoms in 0.2 moles of Na2CO3.10H2O = 13 X 0.2 = 2.6
Now, one mole of a substance contains 6.022 X 10^23 particles of the substance. Thus
number of atoms of oxygen in one mole of oxygen atom = 6.022 X 10^23
number of moles of oxygen atoms in 2.6 moles of oxygen atoms = 2.6 X 6.022 X 10^23 = 15.657 X 10^23
= 1.566 X 10^24
Thus, there are 1.566 X 10^24 atoms of oxygen in 0.2 moles of Na2CO3.10H2O.
4.7
