20 ml of a 1.0 m hcl solution are added to 3.0 l of acid solution with a ph of 5.35. the ph of the mixture rises to 5.33 was the
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<h3>Answer:</h3>
Temperature is 529.164 K
<h3>Explanation:</h3>We are given
Number of moles of Ne (n) = 0.019135 moles
Volume (V) = 878.3 mL
Pressure (P) = 0.946 atm
We are required to calculate the temperature;
We can do this using the ideal gas law equation which is;
PV = nRT, where P is the pressure, n is the number of moles, V is the volume, R is the ideal gas constant (0.082057 Latm/mol/K) and T is the temperature.
From the equation;
Therefore, the temperature will be 529.164 K.
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