Answer:
The gas that Dr. Brightguy added was O₂
Explanation:
Ideal Gases Law to solve this:
P . V = n . R . T
Firstly, let's convert 736 Torr in atm
736 Torr is atmospheric pressure = 1 atm
20°C = 273 + 20 = 293 T°K
125 mL = 0.125L
0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K
(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n
5.20x10⁻³ mol = n
mass / mol = molar mass
0.1727 g / 5.20x10⁻³ mol = 33.2 g/m
This molar mass corresponds nearly to O₂
A) 120 mm
B) 127 mm
C) 914.4 mm
D) 1000 mm
E) 3048 mm
I legitimately think it's 87.3 grams
Answer:
Zn 3
s 1
o 4
Zn. zine
s. sulphate
o oxygen
3znso4. zine sulphateoxide
The answer is A. It is an alkyn. There is a triple bond
