Molarity of a solution if 124.86 g of rbf are dissolved into a solution of water that has a final volume of 2.00L is 0.59.
<h3>What is molarity?</h3>
Molarity is used for dilute aqueous solutions held at a constant temperature. In general, the difference between molarity and molality for aqueous solutions near room temperature is very small and it won't really matter whether you use a molar or molal concentration.
MOLARITY = no of moles of solute/volume of soln in litres
No of moles of rbf = 124.6/104.46
= 1.19
Volume of soln = 2
Molarity=1.19/2 = 0.59
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Answer:
The solution will not form a precipitate.
Explanation:
The Ksp of PbI₂ is:
PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)
Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>
When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:
[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>
[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>
<em />
Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>
If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.
Replacing:
Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹
As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.
Answer:
You can't ice skate on a liquid, when it is frozen it is a solid, when it's unfrozen it is a liquid
Answer:
6× 10⁸
Explanation:
We need to find the multiplication of 2 x 10⁴ by 3 x 10⁴.
2 x 10⁴ × 3 x 10⁴
= (3 × 2) x 10⁴ x 10⁴
= 6 x 10⁴ x 10⁴
= 6 × 10⁴⁺⁴
= 6× 10⁸
Hence, the required answer is 6× 10⁸.
