Answer:
Correct option is
B
5 liters of CH
4
(g)NO
2
at STP
No. of molecules=
22.4
5
mol=
22.4
5
×N
A
molecules
A) 5ℊ of H
2
(g)
No. of moles=
2
5
mol=
2
5
×N
A
molecules
B) 5l of CH
4
(g)
No. of moles of CH
4
=
22.4
5
mol=
22.4
5
N
A
molecules
C) 5 mol of O
2
=5N
A
O
2
molecules
D) 5×10
23
molecules of CO
2
(g)
Molecules of 5l NO
2
(g) at STP=5l of CH
4
(g) molecules at STP
Therefore, option B is correct.
If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.
An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Given :
The ideal gas equation is given below.
n = PV/RT
n = 86126.25 x 0.0024 / 8.314 x 287
n = 0.622 / molar mass (n = Avogardos number)
Molar mass = 7.18 g
Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g
More about the ideal gas equation link is given below.
brainly.com/question/4147359
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Answer:
solubility of X in water at 17.0 
is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 can be calculated using the information given.
Let's assume solubility of X in water at 17.0 is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = 
= 0.11
Hence solubility of X in water at 17.0 is 0.11 g/mL.
