Answer:
It can be removed by acidic chemicals
Explanation:
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 = ![\frac{[HPO4-2] x [ADP]}{ATP}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO4-2%5D%20x%20%5BADP%5D%7D%7BATP%7D)
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 = ![\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2.1x10%5E%7B-3%7D%5D%20x%20%5B8.4x10%5E%7B-3%7D%5D%20%7D%7B%5B1.2%20x%2010%5E%7B-2%7D%5D%20%7D)
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:
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

Thus, by combining them, we obtain:

Which is related to the general line equation:

Whereas:

It means that we answer to the blanks as follows:
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.
Regards!
The freezing point of the sucrose solution is -0.435°C.
<h3>What is the freezing point of the solution?</h3>
The freezing point of the solution is determined from the freezing point depression formula below:
Kf(H₂O) = 1.86 Cm
m is molality of solution = moles of solute/mass of solvent
moles of sucrose = 8.0/342.3 = 0.0233 moles
m = 0.0233/0.1 = 0.233 molal
ΔT = 0.233 m * 1.86°C/m.
ΔT = 0.435 °C.
Freezing point of sucrose solution = 0°C - 0.435°C
Freezing point of sucrose solution = -0.435°C.
In conclusion, the freezing point of sucrose solution is determined from the freezing point depression.
Learn more about freezing point depression at: brainly.com/question/19340523
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