The answer you need is Volume.
6 is the fourth significant figures
if the number behind it is 5 or more then 5, you must add 1 to the number and ALL the number behind it will turn into 0
so that the answer is 42560
I think it’s B not rlly sure
Answer:- partial pressure of Kr = 0.306 atm, partial pressure of oxygen = 0.264 atm and partial pressure of carbon dioxide = 0.396 atm
Total pressure is 0.966 atm
Solution:- moles of Kr = 21.7 g x (1mol/83.8g) = 0.259 mol
moles of oxygen = 7.18 g x (1mol/32g) = 0.224 mol
moles of carbon dioxide = 14.8 g x (1mol/44g) = 0.336 mol
Volume of container = 23.1 L and the temperature is 59 + 273 = 332 K
From ideal gas law equation, P = nRT/V
partial pressure of Kr = (0.259 x 0.0821 x 332).23.1 = 0.306 atm
partial pressure of oxygen = (0.224 x 0.0821 x 332)/23.1 = 0.264 atm
partial pressure of carbon dioxide = (0.336 x 0.0821 x 332)/23.1 = 0.396 atm
Total pressure of the gas mixture = 0.306 atm + 0.264 atm + 0.396 atm = 0.966 atm
Answer:
W = - 10.5943 KJ, work is negative because it is carried out by the system towards the surroundings.
Explanation:
heat engine:
∴ Th = 485°C
∴ Tc = 42°C
∴ Qh = 9.75 KJ......heat from hot source
first law:
∴ ΔU = Q + W = 0 .........cyclic process
⇒ Q = - W
∴ Q = Qh + Qc = Qh - (- Qc)
∴ Qc: heat from the cold source ( - )
⇒ Qh - ( - Qc) = - W..............(1)
⇒ Qc/Qh = - Tc/Th...........(2)
from (2):
⇒ Qc = - (Tc/Th)(Qh) = - (42°C/485°C)(9.75 KJ)
⇒ Qc = - 0.8443 KJ
replacing in (1):
⇒ - W = 9.75 KJ - ( - 0.8443 KJ)
⇒ - W = 10.5943 KJ
