<u>Answer:</u> The volume of oxygen gas required is 3.75 mL
<u>Explanation:</u>
STP conditions:
1 mole of a gas occupies 22.4 L of volume.
We are given:
Volume of ammonia reacted = 3.00 mL = 0.003 L (Conversion factor: 1 L = 1000 mL)
The chemical equation for the reaction of ammonia with oxygen follows:

By Stoichiometry of the reaction:
(4 × 22.4) L of ammonia reacts with (5 × 22.4) L of oxygen gas
So, 0.003 L of ammonia will react with =
of oxygen gas
Hence, the volume of oxygen gas required is 3.75 mL
E=hc/λ =6.626×10^-34×3 ×10^8 / 3×10^7 × 10^-9 = 6.626×10 ^-24J.
Answer:
1.18 moles of gas
Explanation:
3.00 moles of gas are pumped into a 1.00L rigid container with a pressure of 1.98 atm. The gas is released from the container until the pressure is 0.78 atm, how many moles of gas remain in the container?
for the sealed rigid container, the pressure is directly proportional to the amount of gas
3.00moles/1.98 atm = ? moles/0.78 atm
? = 3.00 X 0.78/1.98 =1.18 moles of gas
Answer:
6^3 I believe.
Explanation:
If it's wrong it may be 6^2 because you should square it, but I think it's 6^3