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Answer:
.224 M
Explanation:
To begin, we need to find the molar mass of Na2S
To find Molarity, we use the equation M=moles/Liters of solution
We are given grams, so we must divide by the molar mass of Na2S to find moles.
- 8.76g Na2S / 78.05g/mol Na2S= .112 moles
Now we use our Molarity equation:
- .112 moles Na2S / .500L of solution = .224 M
Let's investigate the substances involved in the reaction first. The compound <span>CH3NH3+Cl- is a salt from the weak base CH3NH2 and the strong acid HCl. When this salt is hydrated with water, it will dissociate into CH3NH2Cl and H3O+:
CH3NH3+Cl- + H2O </span>⇒ CH3NH2Cl + H3O+
Nest, let's apply the ICE(Initial-Change-Equilibrium) table where x is denoted as the number of moles used up in the reaction:
CH3NH3+Cl- + H2O ⇒ CH3NH2Cl + H3O+
Initial 0.51 0 0
Change -x +x +x
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Equilibrium 0.51 - x x x
Then, let's find the equilibrium constant of the reaction. Since the reaction is hydrolysis we use KH, which is the ratio of Kw to Ka or Kb. Kw is the equilibrium constant for water hydrolysis which is equal to 1×10⁻¹⁴. Since the salt comes from the weak base, we use Kb. Since pKb = 3.44, then. 3.44 = -log(Kb). Thus, Kb = 3.6307×10⁻⁴
KH = Kw/Kb = (x)(x)/(0.51 - x)
1×10⁻¹⁴/ 3.6307×10⁻⁴ = x²/(0.51-x)
x = 3.748×10⁻⁶
Since x from the ICE table is equal to the equilibrium concentration of H+, we can find the pH of the aqueous solution:
pH = -log(H+) = -log(x)
pH = -log ( 3.748×10⁻⁶)
pH = 5.43
Answer:
C. 500 cm' of 1.0 mol dmº magnesium sulphate solution.
Explanation:
Let us look at each of the solutions individually;
CaCl2 has three particles
K2SO4 has three particles
MgSO4 has two particles
C2H5OH has only one particle
The number of moles of moles in 250 cm of 2.0 mol dm-3 potassium chloride is 250/1000 * 2 = 0.5 moles having two particles
Also; number of moles in 500 cm' of 1.0 mol dm-3 magnesium sulphate solution= 500/1000 * 1 = 0.5 moles having two particles
Hi there.
Alkali metals are known as the group 1 elements.
When we know this, we just have to look to group 1 of the periodic table.
Rb (Rubidium) is in group 1; it is classified as an alkali metal.
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