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brilliants [131]
3 years ago
10

With a 0.5 M solution: How many moles of NaCl would there be in 1,000 ml?

Chemistry
2 answers:
Triss [41]3 years ago
6 0

<u>Answer:</u> The number of moles of sodium chloride in the solution is

<u>Explanation:</u>

Molarity of the solution is defined as the number of moles that are present in one liter of solution.

The equation used to calculate the molarity of the solution follows:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}

We are given:

Volume of solution = 1000 mL = 1 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.50 moles/ L

Putting values in above equation, we get:

0.5mol/L=\frac{\text{Moles of solute}}{1L}\\\\\text{Moles of the solute}=0.5mol

Hence, the number of moles of NaCl in solution is 0.5 moles.

Shalnov [3]3 years ago
5 0
0.5 moles, though not a verified answer hope it helps! :D
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Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

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4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

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(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

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Answer:

B. only particle Z

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Answer:

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Hope this helped!

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