Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT. Substituting a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
Moles Cu+2 = M * V
= 0.05 L * 0.011 m
= 0.00055 moles
when the molar ratio of Cu2+: EDTA = 1:1 so moles od EDTa also =0.00055 moles
and when the Molarity of EDTa = 0.0630 M
∴ Volume of EDTA = moles / Molarity
= 0.00055 / 0.0630
= 0.0087 L = 8.7 L
Answer:
The new volume of the balloon will be 6046.28 L
Explanation:
Initial pressure (P1) = 99 kpa
initial volume (V1) = 3000 L
Initial temperature = 39 C = 39 + 273 = 312 K
Final pressure (P2) = 45.5 kpa
Final temperature = 16 C = 16 +273 = 289K
Final volume = ????
To calculate the final volume using the general gas equation
P1 V1 / T1 = P2 V2 / T2
make V2 the subject of the formular
V2 = 99000 ×3000× 289 / 45500×312
V2 = 85833000 /14196
V2 = 6046.28 litres
Answer:1.
1.Balanced equation
C4H10 + 9 02 ==> 5H20 +4CO2
2. Volume of CO2 =596L
Explanation:
1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;
CxHy +( x+y/4) O2 ==> y/2 02 + xCO2
Where x and y are number of carbon and hydrogen atoms respectively.
For butane (C4H10)
x=4 and y=10
Therefore
C4H10 + 9 02 ==> 5H20 +4CO2
2. Mass of butane = 0.360kg
Molar mass of C4H10 = ( 12×4 + 1×10)
= 48 +10=58g/mol= 0.058kg/mol
Mole = mass/molar mass
Mole = 0.360/0.058= 6.2moles
From the stoichiometric equation
1mole of C4H10 will gives 4moles of CO2
Therefore
6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2
Using the ideal gas equation
PV=nRT
P= 1.0atm
V=?
n= 24.8mol.
R=0.08206atmL/molK
T=20+273=293
V= 24.8 × 0.08206 × 293
V= 596L
Therefore the volume of CO2 produced is 596L
