If the titrant has a molarity of 0.1000 m and there are 45.00 ml of analyte present, what is the molarity of the analyte?
1 answer:
You might be interested in
<u>larger is the value of </u><u> the higher will the solubility of solid in water.</u>
What is called compound?
- In chemistry, a compound is a substance made up of two or more different chemical elements combined in a fixed ratio.
- When the elements come together, they react with each other and form chemical bonds that are difficult to break.
- These bonds form as a result of sharing or exchanging electrons between atoms.
The equation for the dissociation of a solid MX in water is given below
MX(s) ⇄ Mⁿ⁺(aq) + Xⁿ⁻ (aq)
Assume s be the solubility of MX in pure water, then the equilibrium concentrations of ions are
[ Mⁿ⁺] = s
[ Xⁿ⁻ ] = s
The expression for the solubility product constant () is as follows
= [ Mⁿ⁺] [ Xⁿ⁻ ]
= s²
That is, larger is the value of the higher will the solubility of solid in water.
Learn more about compound
#SPJ4
The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g
First, we will write the balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂
This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate
Now, we will calculate the number of moles of barium hydroxide present.
Mass of barium hydroxide (Ba(OH)₂) = 20 g
Using the formula
Molar mass of Ba(OH)₂ = 171.34 g/mol
∴ Number of moles of Ba(OH)₂ present =
Number of moles of Ba(OH)₂ present = 0.116727 mole
Now,
Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate
Then,
0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate
2 × 0.116727 = 0.233454 mole
∴ Number of moles of NH₄NO₃ required is 0.233454 mole
Now, for the mass of ammonium nitrate (NH₄NO₃) required
From the formula
Mass = Number of moles × Molar mass
Molar mass of NH₄NO₃ = 80.043 g/mol
∴ Mass of NH₄NO₃ required = 0.233454 × 80.043
Mass of NH₄NO₃ required = 18.68636 g
Mass of NH₄NO₃ required ≅ 18.7g
Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g
Learn more on determining mass of reactant required here: brainly.com/question/11232389