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snow_tiger [21]
2 years ago
10

Which statement best explains how the solution should be made? Add 49. 6 mL of 0. 250 M sucrose to 400. 0 mL of water to get 400

. 0 mL of 0. 0310 M sucrose. Add 49. 6 mL of 0. 0310 M sucrose to 350. 4 mL of water to get 400. 0 mL of 0. 250 M sucrose. Add 49. 6 mL of 0. 250 M sucrose to 350. 4 mL of water to get 400. 0 mL of 0. 0310 M sucrose.
Chemistry
1 answer:
soldier1979 [14.2K]2 years ago
7 0
The Last one thank you thank you hope this helped!
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Based on the types of elements in the compound are there ionic or covalent? A. KCI B.NO2
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B.

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Consider the formation of the three solutionsshown in the table.
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2 years ago
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Determine the concentrations of mgcl2, mg2 , and cl– in a solution prepared by dissolving 2.75 × 10–4 g mgcl2 in 1.75 l of water
inysia [295]

Answer:


M of MgCl₂ = 1.65 × 10⁻⁶ M


M of Mg²⁺ = 1.65 × 10⁻⁶ M


M of Cl⁻ = 3.30 × 10⁻⁶ M



Explanation:



1) MgCl₂


Molarity = number of moles of solute / volume of solution in liters, M = n / V


n = mass in grams / molar mass


molar mass of MgCl₂ = 24.305 g/mol + 2(35.543 g/mol) = 95.211 g/mol


n = 2.75 × 10⁻⁴ g / 95.211 g/mol = 2.89×10⁻³ moles


⇒ M = n / V = 2.89×10⁻³ moles / 1.75 l = 1.65 × 10⁻⁶ M



2) Mg²⁺ and Cl⁻


Those are the ions in solution.


You assume 100% dissociation of the ionic compound (strong electrolyte).


Then the equation is: MgCl₂ → Mg²⁺ + 2Cl⁻


That means that 1 mol of MgCl₂ produces 1 mol of Mg²⁺ and 2 moles of Cl⁻.


That yields the same molarity concentration of Mg²⁺ , while the molarity concentration of Cl⁻ is the double.



So, the results are:


M of MgCl₂ = 1.65 × 10⁻⁶ M


M of Mg²⁺ = 1.65 × 10⁻⁶ M


M of Cl⁻ = 3.30 × 10⁻⁶ M

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2 years ago
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