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3 years ago

PLS I NEED HELP give me examples of an atom

Chemistry

1 answer:

Ugo [173]3 years ago

4 0

Answer:

Neon (Ne)

Hydrogen (H)

Argon (Ar)

Iron (Fe)

Calcium (Ca)

Deuterium, an isotope of hydrogen that has one proton and one neutron.

Plutonium (Pu)

F-, a fluorine anion.

Explanation:

I got u

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A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan

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The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = $\frac{18.0 g}{60 g/mol}=0.3 mol$

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

$Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}$

Molarity of the urea solution ;

$M=\frac{0.3 mol}{0.200 L}=1.50 M$

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

$m=d\times V=0.95 g/mL\times 200.0 mL=190 g$

$m = 190 g = 190 \times 0.001 kg = 0.19 kg$

(1 g = 0.001 kg)

Molality of the urea solution ;

$Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}$

$m=\frac{0.3 mol}{0.19 kg}=1.58 m$

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

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