First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g
A: The Equator To The Poles
From Charle's law the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant pressure.
Therefore';
V1/T1=V2/T2
Where; V1 = 2.40 l, T1 = 273 +20= 293 K, V2 = 4.80, and T2= ?
2.4/293= 4.8/T2
T2= (4.8×293)/2.4
= 586 K or 313° C
Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
![q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B%5B%5Ctext%20%7Bglucose%206-phosphate%7D%5D%5BADP%5D%7D%7B%5BGlucose%5D%5BATP%5D%7D)

so,
⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
Answer:
The correct answer is option E.
Explanation:
The Gibbs free energy is given by expression:
ΔG = ΔH - TΔS
ΔH = Enthalpy change of the reaction
T = Temperature of the reaction
ΔS = Entropy change
We have :
ΔH = -720.5 kJ/mol = -720500 J/mol (1 kJ = 1000 J)
ΔS = -263.7 J/K
T = 141.0°C = 414.15 K


The Gibb's free energy of the given reaction at 141.0°C is -611.3 kJ/mol.