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Gekata [30.6K]
2 years ago
6

11. Which is the solubility product expression for Na3PO4(s)?

Chemistry
1 answer:
dlinn [17]2 years ago
8 0

Answer: B

Explanation: the 3 in Na3PO4(s) belong to the sodium atom (Na). so in any of these equations, the 3 would have to be with Na.

A- the 3 is along w the PO4, which would make it part of that bond

C- there is no 3 at all for Na in this choice, making it incorrect

D- Again, the 3 is placed on the other half of the bond

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58.5 moles

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How many grams of nicl2∙6h2o will be used to prepare a 0. 0350 m, 500. 0 ml of nicl2 solution?
marin [14]

The mass of NiCl₂•6HO₂ needed to prepare a 0.035 M 500 mL solution of NiCl₂•6HO₂ is 4.165 g

<h3>What is molarity? </h3>

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

<h3>How to determine the mole of NiCl₂•6HO₂</h3>
  • Molarity = 0.035 M
  • Volume = 500 mL = 500 / 1000 = 0.5 L
  • Mole of NiCl₂•6HO₂ =?

Mole = Molarity × Volume

Mole of NiCl₂•6HO₂ = 0.035 × 0.5

Mole of NiCl₂•6HO₂ = 0.0175 mole

<h3>How to determine the mass of NiCl₂•6HO₂</h3>
  • Mole of NiCl₂•6HO₂ = 0.0175 mole
  • Molar mass of NiCl₂•6HO₂ = 238 g/mol
  • Mass of NiCl₂•6HO₂ =?

Mass = mole × molar mass

Mass of NiCl₂•6HO₂ = 0.0175 × 238

Mass of NiCl₂•6HO₂ = 4.165 g

Thus, 4.165 g of NiCl₂•6HO₂ is needed to prepare the solution

Learn more about molarity:

brainly.com/question/15370276

3 0
2 years ago
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