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Leokris [45]
3 years ago
12

Which experiment best shows water’s ability to act as a solvent? raise the temperature of water and record its boiling point. fr

eeze a flexible vessel of water and watch it expand. pour an ionic solid into water and see if it vanishes. place a leaf on top of the surface of water and see if it sinks?
Physics
2 answers:
IrinaVladis [17]3 years ago
8 0
The vanishing of an ionic solid (like table salt) would be an example of acting like a solvent
IceJOKER [234]3 years ago
4 0

Answer: pour an ionic solid into water and see if it vanishes.

Explanation: The solubility of substances is governed by: Like dissolves like, which states that polar compounds are soluble in polar solvents and non polar compounds are soluble in non polar solvents.

Ionic compounds are formed by combination of positively charged cations and negatively charged anions which are held by strong coloumbic forces.When ionic compounds are dissolved in water, the ions are dissociated due to high dielectric constant of water and thus dissolve in water.

Water being a polar solvent is able to dissolve ionic or polar compounds in it.

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One mole of magnesium (6 × 1023 atoms) has a mass of 24 grams, as shown in the periodic table on the inside front cover of the t
natka813 [3]

This question involves the concepts of density, volume, and mass.

The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".

<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>

It is given that:

Mass of one mole = 24 grams

Mass of 6 x 10²³ atoms = 24 grams

Mass of 1 atom = \frac{24\ grams}{6\ x\ 10^{23}\ atoms} = 4 x 10⁻²³ grams

<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>

\rho = \frac{m}{V}\\\\V=\frac{m}{\rho}

where,

  • \rho = density = 1.7 grams/cm³
  • m = mass of single atom = 4 x 10⁻²³ grams
  • V = volume of single atom = ?

Therefore,

V=\frac{4\ x\ 10^{-23}\ grams}{1.7\ grams/cm^3}

V = 2.35 x 10⁻²³ cm³

<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>

The atom is in a spherical shape. Hence, its Volume can be given as follows:

V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}

d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m

Learn more about density here:

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