Liquid is able to flow because the atoms in liquid are energetic and they have enough energy to slide past each other and move around and fill spaces of a container. Hope it helped!!
Answer:
1.503 mM is the concentration of fructose-6-phosphate.
Explanation:
Glucose-6-phosphate ⇄ Fructose-6-phosphate
The equilibrium constant will be given by = 
![K_c=\frac{[\text{Fructose-6-phosphate}]}{[\text{Glucose-6-phosphate}]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5Ctext%7BFructose-6-phosphate%7D%5D%7D%7B%5B%5Ctext%7BGlucose-6-phosphate%7D%5D%7D)
![K_c=\frac{[\text{Fructose-6-phosphate}]}{2.95 mM}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5Ctext%7BFructose-6-phosphate%7D%5D%7D%7B2.95%20mM%7D)
Relation between standard Gibbs free energy and equilibrium constant follows:
where,
= standard Gibbs free energy = 1.67 kJ/mol = 1670 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = 
T = temperature = ![25.0^oC=[273+25.0]K=298.0 K](https://tex.z-dn.net/?f=25.0%5EoC%3D%5B273%2B25.0%5DK%3D298.0%20K)
![1670 J/mol=-8.314 J/mol K\times 298.0 K\times \ln \frac{[\text{Fructose-6-phosphate}]}{2.95 mM}](https://tex.z-dn.net/?f=1670%20J%2Fmol%3D-8.314%20J%2Fmol%20K%5Ctimes%20298.0%20K%5Ctimes%20%5Cln%20%5Cfrac%7B%5B%5Ctext%7BFructose-6-phosphate%7D%5D%7D%7B2.95%20mM%7D)
On solving above equation :
![[\text{Fructose-6-phosphate}]=1.503 mM](https://tex.z-dn.net/?f=%5B%5Ctext%7BFructose-6-phosphate%7D%5D%3D1.503%20mM)
1.503 mM is the concentration of fructose-6-phosphate.
power = work/time
watt = joule/sec
a) power = 0.10 watt = 0.1joules/sec
work = power * time
= 0.10 * 1 =0.1joules
work is northing but energy
therefore energy is 0.1joules
according to planks quantum theory E = nhν where nis the no. of photons ; h is planks constant; ν isfrequency
hence, n = E/hν
or n = Eλ/hc (ν = c/λ)
n = (0.1*700*10^-9)/6.625*10^-34*3*10^8 = 0.3522*10^18photons
b) similar to the above calculation
here E = 1.0joules
n = (1.0*700*10^-9)/6.625*10^-34*3*10^8 = 0.3522*10^17photons
Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
