approximately how many grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution

Answers

Answer 1

Answer:

87.75 grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution.

What is Mole?

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.

A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.

Given,

Volume of NaCl = 500ml

Concentration = 3M

Moles = Concentration × volume

= 3 × 0.5

= 1.5 moles

Mass = moles × molar mass

= 1.5 × 58.5

= 87.75 g

Therefore, 87.75 grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution.

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Answer 2

Answer: 3.00 M means 3.00 mol/L. So if you want 500 mL=0.5 L, it needs 3.00 mol/L * 0.5 L = 1.50 mol. For NaCl, the molar mass is 23+35.5=58.5 g/mol. So the amount of NaCl needed is 1.50*58.5=87.75 g. So the answer is 87.75 grams.

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Answers

The bond between Br atoms in an Br2 molecule is concealment and is formed by sharing of two valence electrons.

Final answer:

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Explanation:

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In a covalent bond, atoms share electrons to complete their outer electron shell. In the case of a Br2 molecule, each Br atom has 7 valence electrons and needs 1 more electron to achieve a stable octet arrangement, so they share one pair of electrons to achieve this.

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What is the diffrence between acidic and basic solutions in working with redox reactions?

Answers

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Which compound is an electrolyte?
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Answers

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Determine the number of centiliters in 2.4 x 10^3 milliliters.

Answers

2.4(10^3)

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=2400 milliliters

To centiliters is 2400mL= 240.0000cl.

2.4(10^3) To the power off

=2.4*10^3

=2.4*(10*10*10)

=2400 milliliters

To centiliters is 2400mL= 240.0000cl.

Cl= Centiliters

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Answers

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Explanation:

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