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LuckyWell [14K]
3 years ago
12

The density of water is 1.00 g/ml at 4°c. how many water molecules are present in 2.36 ml of water at this temperature?

Chemistry
1 answer:
spin [16.1K]3 years ago
5 0

Mass of 1 ml of water 1 g as density of water 1.00 g/ml at 4° C. So, mass of 2.36 ml of water \frac{1}{2.36} g= 0.423 g.

Molecular mass of water is 18 g which indicates 18 g water contains 6.023 X 10^{23} number of water molecules. So, 0.423 g of water contains \frac{6.023 X 10^{23} X 0.423 }{18}  g = 0.141 X 10^{23} number of water molecules= 1.41 X 10^{22} number of water molecules.

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If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of
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This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

M=\frac{n}{V}

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

2.60g*\frac{1mol}{102.89g} =0.0253mol

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

M=\frac{0.0253mol}{0.160L}=0.158M

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L

That in milliliters turns out to be:

V=0.211L*\frac{1000mL}{1L}=211mL

Learn more about molarity: brainly.com/question/10053901

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Explanation:

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3 years ago
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1 mole of any particles = 6.02* 10²³ particles

4.5*10²⁵ atoms Ni* 1 mol Ni/6.02*10²³ Ni ≈ 74.75≈ 75 mol Ni
 
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