The ratio of isocitrate/citrate under standard conditions is 0.096 when the value of ΔG° is given by 5.8kj/mol.
We know that isocitrate and citrate are convertible. It is given that the ΔG° is given by 5.8kj/mol. Then we know that,
ΔG° = -RTlnK
lnK = -RT/ΔG°
On substituting the values, we get
lnK = -5.6x10³/(6.314x296)
lnK = -2.34
K = 0.09623
The value of k is 0.09623.
Therefore, the ratio of Isocitrate to citrate is given by,
[Isocitrate]/[Citrate] = 0.09623
The ratio could be however rounded off to 0.096.
Therefore, the ratio is given by 0.096 for the isocitrate/citrate under standard conditions. In this way, the ratio could be computed very easily.
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Answer:
603000 J
Explanation:
The following data were obtained from the question:
Energy required (Q) =...?
Mass (M) = 10000 g
Specific heat capacity (C) = 2.01 J/g°C
Overheating temperature (T2) = 121°C
Working temperature (T1) = 91°C
Change in temperature (ΔT) =.?
Change in temperature (ΔT) =T2 – T1
Change in temperature (ΔT) = 121 – 91
Change in temperature (ΔT) = 30°C
Finally, we shall determine the energe required to overheat the car as follow:
Q = MCΔT
Q = 10000 × 2.01 × 30
Q = 603000 J
Therefore, 603000 J of energy is required to overheat the car.
HCl + NaHCO3 = NaCl + H2O + CO2
We have 0.033*0.2 = 0.0066 mol of Hcl
According to reaction we need the same amount of NaHCO3
M(NaHCO3) = 23+1+12+48=84g/mold
m = 0.0066mol * 84g/mol = 0.5544g
Answer:
newton's 3 laws state:
- an object in motion stays in motion (unless an external force acts on it)
- every action has an opposite but equal reaction
- force= mass x acceleration
Answer:
n = 0.3 mol
Explanation:
Given data:
Volume of gas = 8.0 L
Temperature of gas = 45 °C (45+273 = 318 K)
Pressure of gas = 0.966 atm
Moles of gas present = ?
Ideal gas constant = R = 0.021 atm.L/mol.K
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Now we will put the values:
0.966 atm × 8 L = n × 0.0821 atm.L/mol.K × 318 K
7.728 atm.L = n × 26.12 atm.L/mol
7.728 atm.L / 26.12 atm.L/mol = n
n = 0.3 mol
