In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.
Given
P = 10130.0 kPa
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol
Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.
PV = nRT PV
---- ------ ---> n = --------
RT RT RT
10130.0 kPa x 50 L
n= ---------------------------------------------
8.314 L. kPa/K.mol x 573.15 K
506,500
= ----------------------------
4,765.17 mol K
=106.29 mol Ar
So the moles of argon gas is 106.29 moles
Answer:
The answer is
<h2>219.5 mL</h2>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula

From the question
mass = 4500 g
density = 20.5 g/cm³
We have

We have the final answer as
<h3>219.51 mL</h3>
Hope this helps you
Answer:
The temperature change from the combustion of the glucose is 6.097°C.
Explanation:
Benzoic acid;
Enthaply of combustion of benzoic acid = 3,228 kJ/mol
Mass of benzoic acid = 0.570 g
Moles of benzoic acid = 
Energy released by 0.004667 moles of benzoic acid on combustion:

Heat capacity of the calorimeter = C
Change in temperature of the calorimeter = ΔT = 2.053°C



Glucose:
Enthaply of combustion of glucose= 2,780 kJ/mol.
Mass of glucose=2.900 g
Moles of glucose = 
Energy released by the 0.016097 moles of calorimeter combustion:

Heat capacity of the calorimeter = C (calculated above)
Change in temperature of the calorimeter on combustion of glucose = ΔT'



The temperature change from the combustion of the glucose is 6.097°C.
you forgot the well answers
As the gas cools it condenses and becomes a liquid its atoms also become smaller