Answer: The answer is 68142.4 Pa
Explanation:
Given that the initial properties of the cylindrical tank are :
Volume V1= 0.750m3
Temperature T1= 27C
Pressure P1 =7.5*10^3 Pa= 7500Pa
Final properties of the tank after decrease in volume and increase in temperature :
Volume V2 =0.480m3
Temperature T2 = 157C
Pressure P2 =?
Applying the gas law equation (Charles and Boyle's laws combined)
P1V1/T1 = P2V2/T2
(7500 * 0.750)/27 =( P2 * 0.480)/157
P2 =(7500 * 0.750* 157) / (0.480 *27)
P2 = 883125/12.96
P2 = 68142.4Pa
Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa
Answer: p2 = 1.06p1
Explanation: pressure increases with temperature increase.
According to Gass law
P1/T1 = P2/T2
T1 = 20°c = 20 +273 = 293k
T2 = 40°c = 40 +373 = 313k
Therefore
P2 = P1T2/T1 = 313P2/293
P2 = 1.06P1
In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa
V should be in cubic meter:
16L = 0.016 m3
R =

= constant

=

==> P1 * V1 = P2 * V2
V2 =

=
V2 = 0.00581 m3 = 5.81 L
Answer:
See explanation below
Explanation:
In this case we have the starting reactant which is the ethine, In the first step reacts with NaNH₂, a strong base. This base will substract the hydrogen from one of the carbon of the ethine, and form a carbanion. This will react with the propane bromide, displacing the bromine and forming a 5 carbon chain with the triple bond on the carbon 1 and 2.
In the second step, reacts with the lindlar catalyst to do a reduction, and form a double bond between carbon 1 and 2. In essence, compound A is similar to compound B.
Finally B reacts with water in acid and makes a addition reaction, and form an alcohol.
The whole process can be seen in the picture below.
Hope this helps