Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
Explanation:
Pressure in the submarine = 108.9 kPa
Volume, V = 2.4 * 10^5 L
Pressure, P = 116k Pa
Temperature, T = 312 K
Ideal gas law: PV = nRT or n = PV / RT
So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K
= 1.073 *10^4 mol
when temperature is changed to 293K,
PV = nRT or P = nRT / V
=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L
=108.9 K Pa
Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
Answer:
750mmHg
Explanation:
The following data were obtained from the question:
T1 = 127°C = 127 +273 = 400K
T2 = 27°C = 27 +273 = 300K
P1 = 1000mmHg
P2 =?
P1/T1 = P2/T2
1000/400 = P2 /300
Cross multiply
400 x P2 = 1000 x 300
Divide both side by 400
P2 = (1000 x 300)/400
P2 = 750mmHg
Therefore, the new pressure after cooling is 750mmHg
Answer:
the individual atom in the molecule
Explanation:
In chemistry, the ball-and-stick model is a molecular model of a chemical substance. Invidual spheres there represent atoms in the molecule. The bigger atomic number the atom has, the larger diameter of the spheres this atom has in this model.
I hope this answer will help you. Have a nice day !
it is a i believe: Bicarbonate and carbonic acid (A)
Answer:
Volume occupied by oxygen gas at 15 degree centigrade is equal to 
centimeter cube
Explanation:
Assuming Pressure is constant.
where T1 and T2 are temperature in Kelvin
Substituting the give values we get-
Volume occupied by oxygen gas at 15 degree centigrade is equal to centimeter cube
