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natima [27]
1 year ago
5

A solution is formed by dissolving 29.2 grams of sodium chloride (NaCl) in 3.0 liters of solution. The molar mass of NaCl is 58.

5 g/mol.What is the molarity of the solution?

Chemistry
1 answer:
astraxan [27]1 year ago
6 0
<h2>Answer:</h2>

0.17M. Option B is correct

<h2>Explanations:</h2>

The formula calculating the molarity of a solution is given as:

M=\frac{n}{v}

where:

n is the moles of sodium chloride

v is the volume of the solution

Given the following parameters;

The volume of the solution = 3.0L

Determine the moles of NaCl

\begin{gathered} \text{Moles of NaCl=}\frac{Mass}{Molar\text{ mass}} \\ \text{Moles of NaCl=}\frac{29.2g}{23+35} \\ \text{Moles of NaCl=}\frac{29.2g}{58.5} \\ \text{Moles of NaCl=}0.499\text{moles} \end{gathered}

Calculate the required molarity

\begin{gathered} \text{Molarity}=\frac{0.499\text{moles}}{3.0L} \\ Molarity\approx0.17M \end{gathered}

Hence the molarity of the solution is 0.17M

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Answer:

This question is incomplete but the completed question is below

Which Of These Species Is Most Likely To Be A Lewis Acid And Is Also Least Likely To Be A Brønsted Acid? (A) NH4⁺ (B) BF₃ (C) H₂O (D) OH⁻

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A lewis acid is a substance that accepts (or is capable of accepting) a pair of electrons. For example BF₃, while a lewis base is a substance that donates (or is capable of donating) a pair of electrons. For example OH⁻.

If we take a look at the boron (B) in BF₃, it has 3 electrons on it's outermost shell, each of which are bonded to flourine and can still accept a pair of electrons (lone pair). <u>This makes it very likely to be a lewis acid</u>.

Bronsted lowry acid is a substance that donates or can donate a proton or H⁺ (for example HCl) while bronsted lowry base is a substance that accepts or can accept a proton or H⁺ (for example NH₃).

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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
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Answer:

pH = 8.0

Explanation:

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35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

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C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

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