2. The temperature of a gas is increased from 125⁰C to 182⁰C inside of a rigid container. The original pressure of the gas was 1
.22atm, what will the pressure of the gas be after the temperature change?
2 answers:
P1/T1 = P2/T2
125⁰C = 398.15 k
182⁰C = 455.15 k
1.22/398.15 = p2/455.15
p2= 1.39atm
the pressure of the gas be after the temperature change is 1.39 atm
Answer:
1.4 atm
Explanation:
According to Gay Lussac’s law, the pressure of gas is inversely proportional t its temperature, when the volume is constant
Initial temperature = 125 + 273 = 398 K
Final temperature = 182 + 273 = 455 K
Initial pressure = 1.22 atm
1.22 atm / 398 K = P₂/ 455 K
P₂ = 
= 1.4 atm
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Magnesium, written as Mg.
Answer:
1.) Atomic No. 19
No. of proton 19
No. of electron 19
symbol 39K19
2.) Atomic no. 26
Mass No. 56
No. of proton 26
Symbol 56Fe26
