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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Answer:
10 L
Explanation:
The equation of the reaction is;
2BaO2 = 2BaO + O2
Number of moles of BaO2 = 114.7 g/169.33 g/mol = 0.677 moles
From the reaction equation;
2 moles of BaO2 yields 1 moles of O2
0.677 moles of BaO2 yields 0.677 * 1/2 = 0.3385 moles of oxygen
Hence;
PV=nRT
V = ?
P = 127400 Pa or 1.257 atm
T = 455 K
n = 0.3385 moles
R = 0.082 atmLmol-1K-1
V = nRT/P
V = 0.3385 * 0.082 * 455/1.257
V= 10 L
The correct answer is D. magnetic bond