You cannot find it if you do not know the information. Was there a molar mass given or do you know if it is asking for the molar mass?
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
Answer:
a) 210 mL of 95% ethanol
b) 75 mL of water
Explanation:
a) We can use the dilution equation to solve this problem. C₁ and V₁ are the concentration and volume of the stock solution, respectively, while C₂ and V₂ are the concentration and volume of the diluted solution.
C₁V₁ = C₂V₂
We want to find V₁, the volume of ethanol stock we need to dilute to prepare the final solution:
V₁ = (C₂V₂) / C₁
The concentrations are represented in percentages. We substitute in the known values to calculate V₁. The units cancel to leave us with units of mL.
V₁ = (C₂V₂) / C₁ = (70%)(285mL) / (95%) = 210 mL
b) The final solution volume is 285 mL and we have added 210 mL of ethanol, so the remaining volume is from the water that we add:
(285 mL - 210 mL) = 75 mL
Answer:
125 divided by 2 = about 62.5
62.5 mph
Explanation:
Answer:
the reducing flame also called the carburizing flame.
Explanation:
because it gets the oxides of the unknown salts
