From the ideal gas law
pv=nRT , n is therefore PV/RT
R is the
R is gas constant =62.364 torr/mol/k
P=500torr
V=4.00l
T=500+273=773k
n={(500 torr x 4.00l)/(62.364 x773k)}=0.041moles
the number of molecules=moles x avorgadro costant that is 6.022x10^23)
6.022 x 10^23) x0.041=2.469 x10^22molecules
The reaction corresponds to the combustion of propane (C3H8). The balanced reaction is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
The reaction enthalpy is given as:
ΔHrxn = ∑nΔH°f(products) - ∑ nΔH°f(reactants)
= [3ΔH°f(CO2(g)) + 4ΔH°f(H2O(g)] - [1ΔH°f(C3H8(g)) + 5ΔH°f(O2(g)]
= [3(-393.5) + 4(-241.8)] - [-103.9 + 5(0)] = -2043.8 kJ
The enthalpy for the combustion of propane is -2044 kJ
Hello!
We have the following data:
v (volume) = ? (in L)
n (number of mols) = 1,5 mol
T (temperature) = 22 ºC
First let's convert the temperature on the Kelvin scale, let's see:
TK = TºC + 273,15
TK = 22 + 273,15
TK = 295,15
P (pressure) = 100 kPa → P = 100000 Pa → P ≈ 0,987 atm
R (gas constant) = 0,082 atm.L / mol.K
<span>We apply the data above to the Clapeyron equation (gas equation), let's see:
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I hope this helps. =)
Answer:
17.3 L
Explanation:
With all other variables held constant, you can find the missing volume using the Charles' Law equation:
V₁ / T₁ = V₂ / T₂
In this equation, "V₁" and "T₁" represent the initial volume and temperature. "V₂" and "T₂" represent the final volume and temperature. Before we can solve, we need to convert Celsius to Kelvin (because we don't want negative numbers or temp. of 0).
V₁ = 20.0 L V₂ = ? L
T₁ = 0.00 °C + 273.15 = 273.15 K T₂ = -36.2 °C + 273.15 = 236.95 K
V₁ / T₁ = V₂ / T₂ <----- Charles' Law equation
20.0 L / 273.15 K = V₂ / 236.95 K <----- Insert values
0.0732 = V₂ / 236.95 K <----- Simplify left side
17.3 L = V₂ <----- Multiply both sides by 236.95
If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure
