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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The mass of oxygen in 4 moles of C2H4O2 is 128g.
<h3>How to calculate mass?</h3>
The mass of a substance can be calculated by multiplying the number of moles of the substance by its molar mass.
Molar mass of C2H4O2 = 12(2) + 1(4) + 16(2) = 60g/mol
mass of C2H4O2 = 60g/mol × 4 moles = 240grams
However, the mass of oxygen in the compound can be calculated as follows:
molar mass of oxygen/molar mass of compound × 240 grams
32g/mol ÷ 60g/mol × 240g
= 128g
Therefore, the mass of oxygen in 4 moles of C2H4O2 is 128g.
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Answer:
<h2>2.5 litres</h2>
Explanation:
The volume can be found by using the formula for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we're finding the final volume
We have
We have the final answer as
<h3>2.5 litres</h3>
Hope this helps you
368.53-183.45= 185.08
185.08-201.01= -15.93
-15.93+325.89= 309.96
Answer: 309.96