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Ludmilka [50]
3 years ago
11

If a crystal lattice structure is so hard, what causes most ionic compounds to be brittle?

Chemistry
2 answers:
KiRa [710]3 years ago
8 0

Explanation:

The cations and anions are locked tightly into place because of the attraction of their opposite charges - as a result, it's difficult to move the ions and the material is very hard.

nasty-shy [4]3 years ago
3 0

Answer:

d

Explanation:

https://chemistry.stackexchange.com/questions/33322/why-are-so-many-ionic-compounds-brittle

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15.24. Calcium hydroxide, also known as slaked lime, is used in industrial processes in which low concentrations of base are req
Ivahew [28]

The concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M

We'll begin by calculating the number of mole of in 0.16 g of Ca(OH)₂. This can be obtained as follow:

Mass of Ca(OH)₂ = 0.16 g

Molar mass of Ca(OH)₂ = 40 + 2[16 + 1] = 74 g/mol

<h3>Mole of Ca(OH)₂ =? </h3>

Mole = mass / molar mass

Mole of Ca(OH)₂ = 0.16 / 74

<h3>Mole of Ca(OH)₂ = 0.00216 mole </h3>

  • Next, we shall determine the molarity of the stock solution of Ca(OH)₂.

Mole of Ca(OH)₂ = 0.00216 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

<h3>Molarity of Ca(OH)₂ =? </h3>

Molarity = mole / Volume

Molarity of Ca(OH)₂ = 0.00216 / 0.1

<h3>Molarity of Ca(OH)₂ = 0.0216 M</h3>

  • Next, we shall determine the molarity of the diluted solution. This can be obtained as follow:

Volume of stock solution (V₁) = 100 mL

Molarity of stock solution (M₁) = 0.0216 M

Volume of diluted solution (V₂) = 250 mL

<h3>Molarity of diluted solution (M₂) =?</h3>

<h3>M₁V₁ = M₂V₂</h3>

0.0216 × 100 = M₂ × 250

2.16 = M₂ × 250

Divide both side by 250

M₂ = 2.16 / 250

<h3>M₂ = 0.00864 M</h3>

Thus, the molarity of the diluted solution is 0.00864 M

  • Finally, we shall determine the concentration of the hydroxide ions, OH¯ in the diluted solution. This can be obtained as follow:

Ca(OH)₂(aq) —> Ca²⁺(aq) + 2OH¯(aq)

From the balanced equation above,

1 mole of Ca(OH)₂ contains 2 moles of OH¯

Therefore,

0.00864 M Ca(OH)₂ will contain =  2 × 0.00864 = 0.01728 M OH¯

Thus, the concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M

Learn more: brainly.com/question/11471182

7 0
3 years ago
What is the formal charge of nitrogen in no2cl
Gre4nikov [31]

Answer:

The formal charge on nitrogen in NO_{2}Cl is +1.

Explanation:

The structure of NO_{2}Cl is as follows.

(In attachment)

Formal\, charge = Valence\, electron - (nonbonding\, electrons + \frac{Bonding\,electrons}{2})

From the structure, Nitrogen has no non bonding electrons. Nitrogen has four bonds and each bond corresponds to 2 electrons. Hence, nitrogen have eight bonding electrons and five valence electrons.

Formal\,charge\,on\,nitrogen = 5-[0+ \frac{8}{2}]= +1

Therefore, The formal charge on nitrogen in NO_{2}Cl is +1.

6 0
3 years ago
When the cations Na+, K+, Rb+, Cs+ are combined with chloride ion in the gas phase to form ion pairs, which pair formation relea
Gemiola [76]

Answer:

NaCl

Explanation:

5 0
3 years ago
Read 2 more answers
What are chemical formulaes?<br> give your own response!!
lbvjy [14]

Answer:

is a way of presenting information about the chemical proportions of atoms

Explanation:

8 0
3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
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