Data:
V1 = 6.7 liter
T1 = 23° = 23 + 273.15 K = 300.15 K
P1 = 0.98 atm
V2 = 2.7 liter
T2 = 125° = 125 + 273.15 K = 398.15 K
P2 = ?
Formula:
Combined law of ideal gases: P1 V1 / T1 = P2 V2 / T2
=> P2 = P1 V1 T2 / (T1 V2)
P2 = 0.98 atm * 6.7 liter * 398.15 K / (300.15K * 2.7 liter)
P2 = 3.22 atm
Answer:
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The oxidation state of Hydrogen in reactant side is 0 and that in product side is +1. Hence the reaction is oxidation.
The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.
Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.
We recognise that
h * 2 = PT - P * h * 20 = 1atm - 0.037atm
= 0.963 atm
1 * h * 2 = Ph * 2V / R * T
= 0.963 atm x 0.520 L / 0.0821 L atm/
molK * 301
= 0.02 mol h2
= 0.02molZn
So 0.02 mol Zn x 65.39 g/mol
= 1.33 g Zn
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