Deposition is the answer you are looking for :)
If I have 0.725 moles of gas at a temperature of 105 K and a pressure of 3.75 atmospheres the volume of the gas 1.66 litres.
Explanation:
Data given:
number of moles of the gas = 0.725
temperature = 105 K
pressure = 3.75 atm
volume of the gas =?
R = 0.08206 Latm/mole Kelvin
Applying the ideal gas law to calculate the volume of the given gas:
PV = nRT
rearranging the equation to calculate volume:
V = 
putting the values in the equation:
V = 
V = 1.66 Litres.
At a temperature of 105 K and pressure of 3.75 atm, 0.725 moles of gas occupy 1.66 litres of volume.
Explanation:
In this reaction, the reactants are Li and N2. The product is Li3N
So we have;
Li + N2 → Li3N
Upon balancing, we have;
6Li + N2 → 2 Li3N
The sum of the coefficients is 6 + 1 + 2 = 9
Answer:
The complete aerobic oxidation of glucose is coupled to the synthesis of as many as 36 molecules of ATP
Explanation:
Glycolysis, the initial stage of glucose metabolism, takes place in the cytosol and does not involve molecular O2. It produces a small amount of ATP and the three-carbon compound pyruvate. In aerobic cells, pyruvate formed in glycolysis is transported into the mitochondria, where it is oxidized by O2 to CO2. Via chemiosmotic coupling, the oxidation of pyruvate in the mitochondria generates the bulk of the ATP produced during the conversion of glucose to CO2. The biochemical pathways that oxidize glucose and fatty acids to CO2 and H2O.
Answer:
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Explanation:
For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.
For a hypothetical reaction:
xA + yB ⇄ zC
The equilibrium constant is :
![Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D%7D%7B%5BC%5D%5E%7Bz%7D%20%7D)
The given reaction involves the decomposition of H2O into H2 and O2

The equilibrium constant is expressed as :
![Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Since Keq = 5.31*10^-10
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)