Answer:- Solubility of the gas at 759 torr is 0.00753 g/L.
Solution:- From given data, 0.00327 g of gas is soluble in 0.376 L of water at 876 torr.
Solubility of gas at 876 torr pressure is = 0.00327g/0.376L = 0.00869 g/L
Solubility of gases is directly proportional to the pressure. It means, grater is the pressure, more is the solubility of gases.
So, 0.00869/876 = X/759
Where, X is the solubility of the gas at 759 torr.
X = 0.00869(759)/876
X = 0.00753 g/L
8Fe + 6O₂ -----> 4Fe₂O₃
If eight atoms of iron react completely with six molecules of oxygen, 4 molecules of iron (III) oxide will be formed.
Answer:
(a) 
(b) 
(c) 
(d) 
Explanation:
Hello,
In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:
(a) 

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:
![Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}](https://tex.z-dn.net/?f=Ksp%3D%5BBa%5E%7B2%2B%7D%5D%5BSeO_4%5E%7B2-%7D%5D%3D%286.7x10%5E%7B-4%7D%5Cfrac%7Bmol%7D%7BL%7D%20%20%20%29%5E2%5C%5C%5C%5CKsp%3D4.50x10%5E%7B-7%7D)
(B) 

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:
![Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}](https://tex.z-dn.net/?f=Ksp%3D%5BBa%5E%7B2%2B%7D%5D%5BBrO_3%5E-%5D%5E2%3D%287.30x10%5E%7B-3%7D%5Cfrac%7Bmol%7D%7BL%7D%29%283.65x10%5E%7B-3%7D%5Cfrac%7Bmol%7D%7BL%7D%29%5E2%5C%5C%5C%5CKsp%3D1.55x10%5E%7B-6%7D)
(C) 

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:
![Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}](https://tex.z-dn.net/?f=Ksp%3D%5BNH_4%5E%2B%5D%5BMg%5E%7B2%2B%7D%5D%5BAsO_4%5E%7B3-%7D%5D%5E2%3D%281.31x10%5E%7B-4%7D%5Cfrac%7Bmol%7D%7BL%7D%29%5E3%5C%5C%5C%5CKsp%3D2.27x10%5E%7B-12%7D)
(D) 

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:
![Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}](https://tex.z-dn.net/?f=Ksp%3D%5BLa%5E%7B3%2B%7D%5D%5E2%5BMoOs%5E%7B-2%7D%5D%5E3%3D%282%2A1.58x10%5E%7B-5%7D%5Cfrac%7Bmol%7D%7BL%7D%29%5E2%283%2A1.58x10%5E%7B-5%7D%5Cfrac%7Bmol%7D%7BL%7D%29%5E3%5C%5C%5C%5CKsp%3D1.05x10%5E%7B-22%7D)
Best regards.
The answer is C. I hope that helps.
Answer: silicon Si, Germanium GE
Explanation: