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:
weaker and longer
Explanation:
Since there are 3 bonds in ethyne in comparision with the 2 bonds of ethyne between carbon atoms, they are attracted more to each other → the bond gets shorter . And since there are one more bond that supports the union → the bond gets stronger
thus the carbon-carbon double bond in ethene is weaker and longer than the carbon-carbon triple bond in ethyne
Answer: Melting point of the naphthalene is 80
.
Explanation:
Melting point of a substance is temperature at which given solid changes its state from solid to liquid. It is also referred as freezing point of the substance.
Freezing point of substance is a temperature at which given substance changes its state from liquid to solid.
Here, the solid piece of naphthalene is heated and remains at 80°C until it is completely melted.
Answer:
400 or 4x102 indicates only onesignificant figure. (To indicate that the trailing zeros are significant a decimal point must be added. 400. has threesignificant digits and is written as 4.00x102 in scientific notation.) Exact numbers have an infinite number of significant digits but they are generally not reported.
