Determine the number of moles and mass of the sample of argon occupying 37.8L at STP
1 answer:
Answer:
The number of moles and mass of the sample of argon occupying 37.8L at STP are 1.6875 moles and 67.41 grams respectively.
Explanation:
The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
Then you can apply the following rule of three: if by definition of STP 22.4 L are occupied by 1 mole of Ar, 37.8 L of Ar are occupied by how many moles?
amount of moles= 1.6875 moles
The atomic weight of Ar is 39.948 g/mol. So the mass of the sample of argon can be calculated as:
1.6875 moles* 39.948 = 67.41 grams
<u><em>The number of moles and mass of the sample of argon occupying 37.8L at STP are 1.6875 moles and 67.41 grams respectively.</em></u>
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Option B
<u>Explanation:</u>
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Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!