Answer : One mole of an ideal gas occupies a volume of 22.4 liters at STP.
Explanation :
As we know that 1 mole of substance occupies 22.4 liter volume of gas at STP conditions.
STP stands for standard temperature and pressure condition.
At STP, pressure is 1 atm and temperature is 273 K.
By using STP conditions, we get the volume of 22.47 liter.
Hnece, the one mole of an ideal gas occupies a volume of 22.4 liters at STP.
Answer:
go to a calculator and see the answer then make the hypothisis which is the answer
Explanation:
magic
America's age = 243.85 years
<h3>Further explanation</h3>
When converting units, list and multiply all the units until you find the unit you want
Time unit conversion
1 year = 365 days
1 day = 24 hours
1 hour = 3600 s
7.69 x 10⁹ seconds to years :

Answer:
= 19
ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J
ΔG° of the reaction under cellular conditions = 10817.46 J
Explanation:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate
So, the equilibrium constant
can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)


= 19
The formula for calculating ΔG° is shown below as:
ΔG° = - RTinK
ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))
ΔG° = 7295.05957 J
ΔG°≅ - 7295.06 J
b)
Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M
the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M
Equilibrium constant
can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)

0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J