Answer: The freezing point of 3.46 gram of a compound X in 160 gram of benzene is 
Explanation:
The relation of density and molar mass is:
where
d = density = 3.27 g/ L
P = pressure of the gas = 773 torr = 1.02 atm (760 torr = 1atm)
M = molar mass of the gas = ?
T = temperature of the gas = 
R = gas constant = 
The relation of depression in freezing point with molality:
= depression in freezing point = 
= 
= freezing point constant = 5.1
m = molality = 
Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is
The answer would be D, abandoning young before eggs hatched, due to the fact that the eggs would probably fail to develop correctly as a result of improper care. The other choices would actually increase their chances of survival.
the ideal gas equation is PV=nRT
where P=pressure
V=Volume
n=no. of moles
R=universal gas constant
T=temperature
The universal gas constant (R) is 0.0821 L*atm/mol*K
a pressure of 746 mmhg =0.98 atm= 1 atm (approx)
T=37 degrees Celsius =37+273=310 K (convert it to Kelvin by adding 273)
V=0.7 L (only getting oxygen, get 21% of 3.3L)
Solution:
(1 atm)(0.7 L)=n(0.0821 L*atm/mol*K)(310 K)
0.7 L*atm=n(25.451 L*atm/mol)
n=0.0275 mole
Answer:
n=0.0275 mole of oxygen in the lungs.
I think the answer is A but I could be wrong
