Answer:
Molar mass of the gas = 15.15 g/mol
Explanation:
PV = nRT
Where,
P = pressure
n = No. of moles
R = Gas constant
T = Temperature
P = 698 torr, 1 torr = 0.00131579 atm
![698\;torr = 698\times0.00131579 = 0.9184\;atm](https://tex.z-dn.net/?f=698%5C%3Btorr%20%3D%20698%5Ctimes0.00131579%20%3D%200.9184%5C%3Batm)
Temperature = 111 °C = 100 + 273.15 = 384.15 K
V = 48.7 L
R = 0.082057 L atm/mol K
Now, PV = nRT
![n = \frac{PV}{RT}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BPV%7D%7BRT%7D)
![n = \frac{0.9184\times48.7}{0.082057\times384.15}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B0.9184%5Ctimes48.7%7D%7B0.082057%5Ctimes384.15%7D)
=1.4189 mol
Molar mass = Mass/ No. of moles
= 21.5/1.4189
=15.15 g/mol
Answer:
i believe the answer to your question is parallax or parsecs. im sorry im not very specific in this!
Explanation:
The answeris c, chemicalchange because the atoms in woodand oxygen are rearranged.
The ions formed are NH4(+) and S(2-)
The dissolution reaction of (NH4) 2S in water is as follows:
(NH4) 2S ==> 2 NH4 (+) + S (2-).
Ammonium sulfide is the ammonium salt of hydrogen sulfide. It has the formula (NH4) 2S and belongs to the sulfide family.
It is a relatively unstable compound (crystals decomposing at -18 ° C, but exists and is more stable in aqueous solution.) With a pKa exceeding 15, the hydrosulfide ion cannot be significantly deprotonated by ammonia. Thus, such solutions consist mainly of a mixture of ammonia and hydrosulphide of ammonium, it has a smell, close to that of hydrogen sulfide, and its aqueous solutions can be precisely by emitting H2S.
4 infiltration percolation!! I think! Correct me if I’m wrong