The given dehydration equation is,

Cadmiumnitrate tetrahydrate when heated dehydrates releasing the combined water as water vapor. The reaction produces 4 moles of gaseous product water vapor. So, the degree of disorder or randomness increases. Hence, the sign of change in entropy is positive.
This reaction is spontaneous at room temperature even if it is endothermic as the sign of change in entropy is positive.
If we write the equation of the reaction that will take place, it is:
2HNO₃ + Na₂CO₃ → 2NaNO₃ + H₂CO3
The molar ratio of 2HNO₃ : Na₂CO₃ = 1 : 2
Therefore, we can set up the equation:
M₁V₁ = 2M₂V₂
Where the left side of the equation has the molarity and volume of HNO₃ and the right side has the molarity and concentration of Na₂CO₃. Substituting:
M₁ = (2 x 0.108 x 35.7) / 25
M₁ = 0.308 M
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
<h3>
Answer:</h3>
0.95 atm
<h3>
Explanation:</h3>
We are given;
Initial pressure, P1 = 1.0 atm
Initial temperature, T1 =298 K (25°C + 273)
Initial volume, V1 = 0.985 L
Final temperature, T2 = 295 K (22°C + 273)
Final volume, V2 = 1.030 L
We are required to find final air pressure;
Using the combined gas law;

To get, P2 ;



= 0.95 atm
Therefore, the air pressure at the top of the mountain is 0.95 atm
Answer:233 Joules/K
Explanation:
∆H= 26.5KJmol-1
Kelvin temperature = 34.6 + 273 = 307.6 K
No of moles= 2.7 moles
2.70 mole x 26.5 kJ/mole = 71.55kilojoules
∆S=71.55 kilojoules / 307.6 K = 0.233 kilojoules/ K
Convert to JK-1
0.233 kilojoules/ K x 1000 Joules/kilojoule = 233 Joules/K