Saturated had as much solute that it can hold, supersaturated holds more than can normally be dissolved.
Nonrenewable we only use them and renewable we can use them over and over.
Answer:
5.0 moles of water per one mole of anhydrate
Explanation:
To solve this question we must find the moles of the anhydrate. The difference in mass between the dry and the anhydrate gives the mass of water. Thus, we can find the moles of water and the moles of water per mole of anhydrate:
<em>Moles Anhydrate:</em>
7.58g * (1mol / 84.32g) = 0.0899 moles XCO3
<em>Moles water:</em>
15.67g - 7.58g = 8.09g * (1mol / 18.01g) = 0.449 moles H2O
Moles of water per mole of anhydrate:
0.449 moles H2O / 0.0899 moles XCO3 =
5.0 moles of water per one mole of anhydrate
Answer: Concentration of N₂ is 4.8.
M.
Explanation: 
is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For
N2(g) + 2 O2(g) ⇄ 2 NO2(g)
= ![\frac{[NO2]^{2} }{[N2][O2]^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNO2%5D%5E%7B2%7D%20%7D%7B%5BN2%5D%5BO2%5D%5E%7B2%7D%20%7D)
From the question concentration of NO2 is twice of O2:
[NO2] = 2[O2]
Substituting this into :
= ![\frac{[2O2]^{2} }{[N2][O2]^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2O2%5D%5E%7B2%7D%20%7D%7B%5BN2%5D%5BO2%5D%5E%7B2%7D%20%7D)
8.3.
= ![\frac{4O2^{2} }{[N2].O2^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B4O2%5E%7B2%7D%20%7D%7B%5BN2%5D.O2%5E%7B2%7D%20%7D)
[N2] = 
[N2] = 
[N2] = 4.8.
The concentration of N2 in the equilibrium is [N2] = 4.8.M.
The answer is roughly 165 cups of beer.
