The chemical reaction is expressed as:
<span> 4KO2+2CO2→2K2CO3+3O2
</span>
We are given the amount of KO2 to be used in the reaction. This will be the starting point of our calculations. We do as follows:
0.425 mol KO2 ( 3 mol O2 / 4 mol KO2 ) = 0.319 mol O2 produced
Hope this answers the question. Have a nice day.
Answer:
THE NEW VOLUME OF THE OXYGEN GAS AT 28 PSI FROM 72.5 PSI IS 0.078 L.
Explanation:
Initial volume of the oxygen in the container = 30.0 mL = 30 / 000 L = 0.03 L
Initial pressure of the oxygen = 72.5 psi = 1 psi = 6890 pascal
Final pressure = 28 psi
Final volume = unknown
First convert the mL to L and since both pressures are in similar unit that is psi; there is no need converting them to pascal or other standard unit of pressure. They cancel each other out.
This question follows Boyle's equation of gas laws and mathematically it is written as:
P1 V1 = P2 V2
Re-arranging by making P2 the subject of the formula, we have:
V2 = P1 V1 / P2
V2 = 72,5 * 0.03 / 28
V2 = 2.175 /28
V2 = 0.0776 L
The new volume of the oxygen gas at a change in pressure from 72.5 psi to 28 psi is 0.078 L.
Answer:
Final volume=V₂ = 216.3 mL
Explanation:
Given data:
Initial volume = 120.0 mL
Initial temperature = -12.3 °C (-12.3 +273 = 260.7 K)
Final volume = ?
Final temperature = 197.0 °C (197+273 = 470 K)
Solution:
We will apply Charles Law to solve the problem.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 120 mL × 470 K /260.7K
V₂ = 56400 mL.K /260.7K
V₂ = 216.3 mL
Yeah, a lot faster than sound.
Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
mass = 0.508 g, Volume = 0.175 L
Temperature = (25 + 273) K = 298 K, P = 1 atm
As per the ideal gas law, PV = nRT.
where, n = no. of moles = 
Hence, putting all the given values into the ideal gas equation as follows.
PV = 
1 atm \times 0.175 L = 
= 71.02 g
As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of 
is 70.8 g/mol or 71 g/mol (approx).
Thus, we can conclude that the gas is most likely chlorine.
