If you decreased the volume of a sample of gas by a factor of three while maintaining a constant pressure, how would the absolut
e temperature of the gas be affected? A gas at 300 K and 4.0 atm is moved to a new location with a temperature of 250 K. The volume changes from 5.5 L to 2.0 L. What is the pressure of the gas at the new location? What is the partial pressure of 0.50 mol Ne gas combined with 1.20 mol Kr gas at a final pressure of 730 t
2 answers:
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Answer:
25.76 L
Explanation:
Given, Volume of Copper = 3.56 cm³
Density = 8.95 g/cm³
Considering the expression for density as:
So,
So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g
Mass of copper = 31.862 g
Molar mass of copper = 63.546 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>Moles of copper = 0.5014 moles </u>
Given, Volume of nitric acid solution = 200 mL = 200 cm³
Density = 1.42 g/cm³
Considering the expression for density as:
So,
So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g
Also, Nitric acid is 68.0 % by mass. So,
Mass of nitric acid = = 193.12 g
Molar mass of nitric acid = 63.01 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>Moles of nitric acid = 3.0649 moles </u>
According to the reaction,
1 mole of copper react with 4 moles of nitric acid
Thus,
0.5014 moles of copper react with 4*0.5014 moles of nitric acid
Moles of nitric acid required = 2.0056 moles
Available moles of nitric acid = 3.0649 moles
<u>Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent. </u>
The formation of the product is governed by the limiting reagent. So,
1 mole of copper on reaction forms 2 moles of nitrogen dioxide
So,
0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide
<u>Moles of nitrogen dioxide = 1.0028 moles </u>
Given:
Pressure = 735 torr
The conversion of P(torr) to P(atm) is shown below:
So,
Pressure = 735 / 760 atm = 0.9632 atm
Temperature = 28.2 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.2 + 273.15) K = 301.35 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K
<u>⇒V = 25.76 L</u>