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VMariaS [17]
2 years ago
12

What is the molarity of a solution which contains 0.80 mol of HCl in 250mL of solution?

Chemistry
1 answer:
mariarad [96]2 years ago
6 0

Answer:

3.2M HCl Solution

Explanation:

Molarity = moles of solute / volume of solution expressed in liters

moles of solute = 0.80 moles HCl

volume of solution = 250 ml = 0.250 Liter

Molarity (M) = 0.80 moles HCl / 0.250 Liters = 3.2M HCl Solution

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The correct answer is C :)

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Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc
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Answer:

  • The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
  • The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%  
  • The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = \frac{4 }{3} \pi r^2

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = \frac{a\sqrt{2} }{4}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3 = \frac{\pi *a^3\sqrt{2}}{24}

Number of atoms/unit cell = 4

Total volume of the atoms = 4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6} = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = \frac{a\sqrt{3} }{4}

Volume of each atom =\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3 =\frac{\pi *a^3\sqrt{3}}{16}

Number of atoms/unit cell = 2

Total volume of the atoms = 2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8} = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = \frac{a\sqrt{3} }{8}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3 = \frac{\pi *a^3\sqrt{3}}{128}

Number of atoms/unit cell = 8

Total volume of the atoms = 8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16} = 0.3401

= 0.3401  X 100% = 34.01%

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3 years ago
The halogens, the elements of Family 17 on the periodic table, combine easily with elements from Family 1. When a halogen reacts
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Answer:

When halogen elements react with group one metals they form halide salts.

Explanation:

The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements can not be found free in nature. Their chemical properties are resemble greatly with each other. As we move down the group in periodic table size of halogens increases that's way fluorine is smaller in size as compared to other halogens elements. Their boiling points also increases down the group which changes their physical states. i.e fluorine is gas while iodine is solid.

When halogen elements react with group one metals they form halide salts.

Alkali metals have one valance electron and halogens needed one electron to complete the octet thus alkali metals loses one electron which is accepted by halogens atom and form ionic compound called halide salts.

For example:

2Na + Cl₂ → 2NaCl

2K + Cl₂  → 2KCl

2Rb + Cl₂ → 2RbCl

2Li + Cl₂ → 2LiCl

With bromine:

2Na + Br₂ → 2NaBr

2K + Br₂  → 2KBr

2Rb + Br₂ → 2RbBr

2Li + Br₂ → 2LiBr

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2Na + I₂ → 2NaI

2K +  I₂ → 2KBI

2Rb + I₂ → 2RbI

2Li + I₂ → 2LiI

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Explanation:

What is motion? define friction.

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When looking at the equilibrium between calcium sulfate and its aqueous ions, what could be added to solution to promote precipi
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The common ion effect is the decreased solubility of the ionic precipitate in the solution. To promote the precipitation of the calcium sulfate, magnesium sulfate must be added.

<h3>What is the common ion effect?</h3>

It is a phenomenon that is due to the addition of the common ion already present in the solution that affects the equilibrium and is used to decrease the solubility of the solute present in the solution to increase the precipitation rate.

Magnesium sulfate will be added to the solution as it already contains the sulfate ion in the form of calcium sulfate. This will increase the precipitation of the compound as the common ion added will decrease the solubility.

Therefore, magnesium sulfate will be added to promote precipitation.

Learn more about common ion effect here:

brainly.com/question/17278748

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