Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols
For every 1 mol of gas, there will be
24000 cm^3 of gas
Vol. = 1.14 x 24 dm^3
= 27.36 dm^3
Please correct me if I'm wrong but I think the answer is b or c
also sorry if i do get it wrong
Frequency = speed of light
---------------------------
wavelength
= 3 x 10^8
------------------------
344 x 10^-9
= 8.72 x 10^14 Hz.
Hope this helps!
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
Answer:
43.05 moles of Al needed to react with 28.7 moles of FeO.
Explanation:
Given data:
Moles of FeO = 28.7 mol
Moles of Al needed to react with FeO = ?
Solution:
Chemical equation:
2Al + 3FeO → 3Fe + Al₂O₃
Now we will compare the moles of Al with FeO.
FeO : Al
2 : 3
28.7 : 3/2×28.7 = 43.05 mol
Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.
