Answer:
m = 72.6 grams
Explanation:
Given that,
The density of the substance uis 1.2 g/mL
Volume, V = 60.5 mL
We need to find the mass of the substance. Density is equal to mass divided by its volume. So, it can be given by :

So, the mass of the substance is 72.6 grams.
It is A because it has a lower activation emery than like B
Answer:
The answer to your question is: yield = 56.27%
Explanation:
Data
CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br
18.54 ml 1-butanol 15.65 g of 1-bromobutane
% yield = ?
density = 0.81 g/ml
MM = 74 g 1- butanol
MM = 137 g 1-bromobutane
Process
Calculate mass of 1- butanol
density = mass/volume
mass = density x volume
mass = 0.81 x 18.54
mass = 15.02 g of 1-butanol
Theoretical yield
74 g of 1- butanol ----------------- 137 g of 1-bromobutane
15.02 g of 1- butanol ------------- x
x = (15.02 x 137) / 74
x = 27.81 g of 1-bromobutane
% yield = experimental yield / theoretical yield x 100
% yield = 15.65 / 27.81 x 100
% yield = 56.28
Answer:
the pressure of the gas in torr is 2280 torr
Explanation:
The pressure of a gas in torr can be calculated by converting its pressure in kPa to atmospheres and then converting the atmospheric pressure to torr.
One atmosphere is equal to 101.325 kPa. Therefore, the pressure of the gas in atmospheres is 305 kPa / 101.325 kPa/atm = 3.00 atm.
One torr is equal to 1/760 of an atmosphere. Therefore, the pressure of the gas in torr is 3.00 atm * 760 torr/atm = 2280 torr.
Therefore, the pressure of the gas in torr is 2280 torr