8. The following system contains 20.0 psi each of gases A and B in the end bulbs with an evacuated bulb in the centre. Both stop
cocks are opened at once. What is the new overall pressure in the system?
I literally have no idea how to even start this, I know that the answer is 13.3 psi, but I don't have any clue why or how to get there.
I literally have no idea how to even start this, I know that the answer is 13.3 psi, but I don't have any clue why or how to get there.
1 answer:
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Answer:
8.34
Explanation:
1) how much moles of NH₃ are in the reaction;
2) how much moles of H₂ are in the reaction;
3) the required mass of the H₂.
all the details are in the attachment; the answer is marked with red colour.
Note1: M(NH₃) - molar mass of the NH₃, constant; M(H₂) - the molar mass of the H₂, constant; ν(NH₃) - quantity of NH₃; ν(H₂) - quantity of H₂.
Note2: the suggested solution is not the shortest one.
Answer: The empirical formula for the given compound is
Explanation:
The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:
where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.
We are given:
Mass of
Mass of
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 21363 g of carbon dioxide, of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in 6125 g of water, of hydrogen will be contained.
Now we have to calculate the mass of nitrogen.
Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Nitrogen =
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.
For Carbon =
For Hydrogen =
For Nitrogen =
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 5 : 7 : 1
Hence, the empirical formula for the given compound nicotine is