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sergey [27]
3 years ago
14

The radius of a hydrogen atom is 37 pm and its mass is 1.67 × 10-24 g. What is the density of the atom in grams per cubic centim

eter? Assume the atom is a sphere with:
Volume=(4/3)(3.14)(r^3)
Chemistry
2 answers:
RoseWind [281]3 years ago
4 0

Answer:

V = 4/3 * 3.1416 * (37x10-10)3

V = 2.12x10-25 cm3

d = m/V

d = 1.67x10-24 / 2.12x10-25 = 7.87 g/cm3

The difference in temperature, let's convert F to ºC:

ºC = -80-32/1.8 = -62.22 ºC

dT = -92.6 + 62.2 = -30.4 ºC

charle [14.2K]3 years ago
3 0

Answer:

V = 4/3 * 3.1416 * (37x10-10)3

V = 2.12x10-25 cm3

Explanation:

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2 years ago
A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. What is the weight/
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Answer:

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :

<u>2.67%</u>

Explanation:

Note : Look at the density of potassium nitrate in water if given in the question.

<u><em>You are calculating </em></u><u><em>weight /Volume</em></u><u><em> not weight/weight % or percent by mass of the solute</em></u>

Here the <u>weight/weight % or percent by mass</u> of the solute is asked : So first convert the<u> VOLUME OF SOLUTION into MASS</u>

Density of potassium nitrate in water KNO3 = 2.11 g/mL

density=\frac{mass}{volume}

Density = 2.11 g/mL

Volume of solution = 86.4 mL

2.11=\frac{mass}{86.4}

mass = 2.11\times 86.4

mass=182.3grams

Mass of Solute = 4.87 g

Mass of Solution = 183.2 g

w/w% of the solute =

= \frac{mass\ of\ solute}{mass\ of\ solution}\times 100

=\frac{4.87}{183.2}\times 100

w/w%=2.67%

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What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent
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The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

<u>Explanation:</u>

Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

         \text {Molality}=\frac{\text {Moles of solute}}{\text {Mass of solvent}}

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.

Hence, \text { molality }=\frac{1.34}{2.47}= 0.54 \mathrm{M}

Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

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