Under which conditions of temperature and pressure would helium behave most like an ideal gas?
2 answers:
Answer : The correct option is, (3) 750 K and 20 kPa
Explanation :
The conditions for ideal gas are :
Ideal gas are those gas that has no intermolecular attractions.
Ideal gas are those gas that have negligible volume.
The ideal gas equation is,
The conditions for real gas are :
Real gas are those gas that have intermolecular attractions.
Real gas are those gas that have volume.
The real gas equation is,
A real gas behave ideally at high temperature and low pressure condition.
From the given options, option (3) have high temperature and low pressure is the correct option.
Hence, at 750 K and 20 kPa conditions of temperature and pressure does a sample of helium behave most like an ideal gas.
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Answer:
(B) 0.160 M
Explanation:
Considering:
Or,
Given :
For :
Molarity = 0.200 M
Volume = 20.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 20.0×10⁻³ L
Thus, moles of :
Moles of = 0.004 moles
For :
Molarity = 0.400 M
Volume = 30.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume =30.0×10⁻³ L
Thus, moles of :
Moles of = 0.012 moles
According to the given reaction:
1 mole of potassium carbonate react with 1 mole of barium nitrate
0.004 moles potassium carbonate react with 0.004 mole of barium nitrate
Moles of barium nitrate = 0.004 moles
Available moles of barium nitrate = 0.012 moles
Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of potassium carbonate gives 1 mole of barium carbonate
Also,
0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate
Mole of barium carbonate = 0.004 moles
Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)
Left over moles = 0.012 - 0.004 moles = 0.008 moles
Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L
So, Concentration = 0.008/0.05 M = 0.160 M
<u>(B) is correct.</u>