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Ludmilka [50]
2 years ago
11

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

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

Answer:

d

Explanation:

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

You might be interested in
Hydrogen gas at a pressure of 740. mmHg has a volume of 2.00 L at a temperature of 25.0°C. What is the temperature of this gas a
nikdorinn [45]

Answer:

The final temperature of hydrogen gas is  537.63 K.

Explanation:

Given data:

Initial volume = 2.00 L

Initial pressure = 740 mmHg (740/760 = 0.97 atm)

Initial temperature = 25 °C (25 +273 = 298 K)

Final temperature =?

Final volume = 3.50 L

Final pressure = standard = 1 atm

Formula:  

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂

T₂  =  P₂V₂T₁  / P₁V₁

T₂ = 1 atm × 3.5 L × 298 K / 0.97 atm × 2.00 L  

T₂ = 1043  atm .L. K / 1.94 atm. L

T₂ = 537.63 K

7 0
3 years ago
What happens to the temperature of the remaining liquid when some of the liquid evaporates
Dovator [93]
Turns into vapor. not all of the molecules are liquid have the same energy
7 0
3 years ago
Which of these solutions is the most basic?
dalvyx [7]
<h2>Question↷</h2>

Which of these solutions is the most basic?

[OH-] = 4.3 x 10-9 M

[OH-] = 2.8 x 10-11 M

[OH-] = 6.7 x 10-10 M

[OH-] = 1.0 x 10-5 M

<h2>Answer↷</h2>

  • <u>[OH-] = 1.0 x 10-5 M</u> ✓

<h2>Solution↷</h2>

[OH-] = 4.3 x 10-9 M

  • [OH-] = 4.3 x 10^-9 M
  • pOH = - log [OH^1-]
  • pOH = - log [4.3 x 10^-9]
  • pOH = 8.36
  • pH = 14- 8.36
  • pH = 5.64

_______________________________________

[OH-] = 2.8 x 10-11 M

  • [OH-] = 2.8 x 10^-11 M
  • pOH = - log [OH^1-]
  • pOH = - log [2.8 x 10^-11 ]
  • pOH = 10.55
  • pH = 14-10.55
  • pH = 3.45

_______________________________________

[OH-] = 6.7 x 10-10 M

  • [OH-] = 6.7 x 10^-10 M
  • pOH = - log [OH^1-]
  • pOH = - log [6.7 x 10^-10]
  • pOH = 9.17
  • pH = 14-9.17
  • pH = 4.83

_______________________________________

[OH-] = 1.0 x 10-5 M

  • [OH-] = 1.0 x 10-5 M
  • pOH = - log [OH^1-]
  • pOH = - log [1.0 x 10^-5 ]
  • pOH = 5
  • pH = 14-5
  • pH = 9

_______________________________________

we know that , the solution with <u>pH > 7</u> is termed as basic and more the pH ,more the basicity, hence ,the solution with the highest pH would be the strongest base out of these all which is

<u>[OH-] = 1.0 x 10-5 M</u> with pH = 9

_______________________________________

4 0
2 years ago
Read 2 more answers
A liquid that occupies a volume of 0.820 L has a mass of 2.56 kg.
user100 [1]

Answer:

The density of this liquid is 0.320 kg/L

Explanation:

Given:

Volume of the Liquid =  0.820 L

Mass of the liquid  = 2.56 kg.

To Find:

The density of the liquid in  kg/L

Solution:

Density is the mass occupied by the substance in unit volume. This density is essential determining whether the substance floats or sinks. Greek letter(rho) is used to denote density

The equation of for density is

Density = \frac{mass}{volume}

\rho = \frac{m}{v}

where m is the mass

v is the volume

On substituting the given values

\rho = \frac{0.820 }{2.56}

\rho = 0.320 kg/L

7 0
3 years ago
A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
sesenic [268]

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

  • 0 mL of HBr is added

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

  • After adding 12 mL of HBr

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

  • Equivalence point

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

  • Titration midpoint (28.2 mL/2)

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

  • When we add 45.1 mL of HBr

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

7 0
3 years ago
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