• 1) The pressure of a sample of gas in a 2.00-L container is 876 mmHg.
1 answer:
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Answer:
1.84 L
Explanation:
Using the equation for reversible work:
Where:
W is the work done (J) = -287 J.
Since the gas did work, therefore W is negative.
P is the pressure in atm = 1.90 atm.
However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)
Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J
is the initial volume = 0.350 L
is the final volume = ?
Thus:
(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( - 0.350) L
= [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L
<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of hex-3-ene = 5.84 g
Molar mass of hex-3-ene = 82.14 g/mol
Putting values in equation 1, we get:
The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:
By Stoichiometry of the reaction:
1 mole of hex-3-ene produces 1 mole of 3-hexanol
So, 0.0711 moles of hex-3-ene will produce = of 3-hexanol
Now, calculating the mass of 3-hexanol from equation 1, we get:
Molar mass of 3-hexanol = 102.2 g/mol
Moles of 3-hexanol = 0.0711 moles
Putting values in equation 1, we get:
To calculate the percentage yield of 3-hexanol, we use the equation:
Experimental yield of 3-hexanol = 4.33 g
Theoretical yield of 3-hexanol = 7.27 g
Putting values in above equation, we get:
Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %