Determine the density of nh3 gas at 435k and 1.00atm
1 answer:
We assume that this gas is ideal. Therefore, we can use the ideal gas equation which is expressed as:
PV=nRT
We manipulate this equation to give us an expression which will correspond to density. We do as follows:
PV= nRT
P/RT = n/V where n = m/MM
P(MM) /RT = m/V = density
Density = 1.00 (17.03) / 0.08206 (435)
Density = 0.48 g / L
You might be interested in
Answer:
mark me as a brainlist answer
Explanation:
Oxygen
Answer:
Because CLEARLY, each mole of glucose, C6H12O6 contains 6⋅mol oxygen atoms.
Answer:
eight
Explanation:
That means there are 12 electrons in a magnesium atom
hope this helps
Answer:
heat increase, pressue loss, altitude gain,
Explanation:
Answer: Option B. 76.83L
Explanation:
1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.
If 1 mole of Radon = 22.4L
Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L