The molar mass of an unknown material is 78.430 g/mol. What is the mass of half of a mole of the unknown material?
2 answers:
You might be interested in
Answer:
ΔG° of reaction = -47.3 x J/mol
Explanation:
As we can see, we have been a particular reaction and Energy values as well.
ΔG° of reaction = -30.5 kJ/mol
Temperature = 37°C.
And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:
The first step is to calculate the equilibrium constant for the reaction:
Equilibrium Constant K =
And we have values given for these quantities in the biological cell:
[HP04-2] = 2.1 x M
[ATP] = 1.2 x M
[ADP] = 8.4 x M
Let's plug in these values in the above equation for equilibrium constant:
K =
K = 1.47 x M
Now, we have to calculate the ΔG° of reaction for the biological cell:
But first we have to convert the temperature in Kelvin scale.
Temp = 37°C
Temp = 37 + 273
Temp = 310 K
ΔG° of reaction = (-30.5 ) + (8.314)x (310K)xln(0.00147)
Where 8.314 = value of Gas Constant
ΔG° of reaction = (-30.5 x ) + (-16810.68)
ΔG° of reaction = -47.3 x J/mol
Answer:
Explanation:
From the given information:
The density of O₂ gas =
here:
P = pressure of the O₂ gas = 310 bar
=
= 305.97 atm
The temperature T = 415 K
The rate R = 0.0821 L.atm/mol.K
molar mass of O₂ gas = 32 g/mol
∴
= 287.37 g/L
To find the density using the Van der Waal equation
Recall that:
the Van der Waal constant for O₂ is:
a = 1.382 bar. L²/mol² &
b = 0.0319 L/mol
The initial step is to determine the volume = Vm
The Van der Waal equation can be represented as:
where;
R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol
Replacing our values into the above equation, we have:
After solving;
V = 0.1152 L
∴
= 277.77 g/L
We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.