Based on the ideal gas relation:
PV = nRT
where P = pressure ; V = volume ; T = temperature
n = number of moles; R = gas constant = 0.0821 L atm/mol-K
Step 1: Find the number of moles of O2
n = PV/RT = 1 * 3.90/0.0821*273 = 0.1740 moles
Step 2: Calculate the molecules of O2
Now, 1 mole of O2 corresponds to 6.023 * 10²³ molecules of O2
Therefore, 0.1740 moles of O2 corresponds to-
0.1740 moles of O2 * 6.023*10²³ molecules of O2/1 mole of O2
= 1.048 * 10²³ molecules of O2
The answer is on the paper
Answer:
NaOH(aq)
Explanation:
NaOH(aq) is known to precipitate Mn^2+ ions according to the following reaction; Mn^2+(aq)+2OH^−(aq)↽−−⇀Mn(OH)2(s)
Hence, manganese(II) oxide reacts more readily with NaOH(aq) under ordinary conditions precipitating the metal hydroxide solid. This is one of the characteristic reactions of Mn^2+.
Answer:
It's true :) Hope that helps
