Pedro’s teacher wrote the formula for glucose, a common sugar, on the board. Pedro learned that the letters in the formula repre
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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:
Hello!
In this case, since the freezing point depression caused by the addition of a solute, we use the following formula:
Thus, we first need to compute the molality of each solute, as shown below:
Next, since NaCl has two ionic species, one Na⁺ and one Cl⁻, and CaCl₂ three, one Ca²⁺ and two Cl⁻, the van't Hoff's factors are 2 and 3 respectively, therefore the freezing point depressions turn out:
It means that CaCl₂ is still better than NaCl because produces involves a higher melting point for the ice, so it would melt faster.
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