Answer:
pH = 5.47
Explanation:
The equilibrium that takes place is:
HIO ↔ H⁺ + IO⁻
Ka = ![\frac{[H+][IO-]}{[HIO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%2B%5D%5BIO-%5D%7D%7B%5BHIO%5D%7D)
= 2.3 * 10⁻¹¹
At equilibrium:
<u>Replacing those values in the equation for Ka and solving for x:</u>
Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47
The balanced chemical reaction is:
<span>Ca + Cl2 = CaCl2
</span>
We are given the amount of calcium metal to be used for this reaction. This will be the starting point for the calculations.
56 g Ca ( 1 mol Ca / 40.08 g Ca) (1 mol Cl2 / 1 mol Ca) ( 22.414 L Cl2 / 1 mol Cl2 ) = 31.32 L Cl2 gas produced from the reaction
Answer:
48.32 g of anhydrous MnSO4.
Explanation:
Equation of dehydration reaction:
MnSO4 •4H2O --> MnSO4 + 4H2O
Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)
= 223 g/mol
Mass of MnSO4 • 4H2O = 71.6 g
Number of moles = mass/molar mass
= 71.6/223
= 0.32 mol.
By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4
Number of moles of MnSO4 = 0.32 mol.
Molar mass = 55 + 32 + (4*16)
= 151 g/mol.
Mass = 151 * 0.32
= 48.32 g of anhydrous MnSO4.
