This compound is also known as Barium Carbonate. 1 mole is equal to 1 moles BaCO3, or 197.3359<span> grams.</span>
Answer:
Ka = 6.02x10⁻⁶
Explanation:
The equilibrium that takes place is:
We <u>calculate [H⁺] from the pH</u>:
- [H⁺] =
Keep in mind that [H⁺]=[A⁻].
As for [HA], we know the acid is 0.66% dissociated, in other words:
We <u>calculate [HA]</u>:
Finally we <u>calculate the Ka</u>:
- Ka =
![\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B9.12x10%5E%7B-4%7D%5D%2A%5B9.12x10%5E%7B-4%7D%5D%7D%7B%5B0.138%5D%7D)
= 6.02x10⁻⁶
Answer:
V₂ = 0.6 V.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n is constant, and have different values of P, V and T:
<em>(P₁V₁T₂) = (P₂V₂T₁).</em>
<em></em>
V₁ = V, P₁ = P, T₁ = T.
V₂ = ??? V, P₂ = 1.25 P, T₂ = 0.75 T.
<em>∴ V₂ = (P₁V₁T₂)/(P₂T₁) =</em> (P)(V)(0.75 T)/(1.25 P)(T)<em> = 0.6 V.</em>
The number of carbon atoms in an alcohol affects its solubility in water, as shown in Table 13.3. As the length of the carbon chain increases, the polar OH group becomes an ever smaller part of the molecule, and the molecule becomes more like a hydrocarbon. The solubility of the alcohol decreases correspondingly.
I would have to say it would be the closest to a gas A.
