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3 years ago

Which characteristic influences the polarity of these molecules?

Chemistry

1 answer:

mash [69]3 years ago

4 0

Answer:

The larger the difference in electronegativity, the larger the dipole moment.

Explanation:

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Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

Explanation :

As we know that CaCO₃ dissociates to give $Ca^{2+}$ ion and $CO_3^{2-}$ ion.

The solubility equilibrium reaction will be:

$CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][CO_3^{2-}]$

Let solubility of CaCO₃ be, 's'

$K_{sp}=(s)\times (s)$

$K_{sp}=s^2$

$8.7\times 10^{-9}=s^2$

$s=9.33\times 10^{-5}mol/L$

Therefore, the solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

4 0

3 years ago

Which of the following is a hypothesis for an experiment measuring solubility of sodium sulfate?

SCORPION-xisa [38]

A would be a hypothesis

7 0

3 years ago

6. A quiet sound exerts a pressure of 4.00 x 10^-8 kPa (kPa= kilopascals, a pressure unit). What is this pressure measured in at

Radda [10]

Answer:

GOOGLE MY HUMAN

Explanation:

GOOGLE!!!!!

3 0

3 years ago

Why the partial charge on H in HF is more positive than the partial charge on H in HBr

erik [133]

Because fluorine has a higher electronegativity

8 0

3 years ago

An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the

san4es73 [151]

Answer:

the ionic radius of the anion $r^- = 312.52 \ pm$

Explanation:

From the diagram shown below :

The anion $Cl^-$ is located at the corners

The cation $Cs^+$ is located at the body center

The Body diagonal length = $\sqrt{3 \ a }$

∴ $2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a$

Given that :

$\frac{r^+}{r^-} =0.84$ (i.e the ratio of the ionic radius of the cation to the ionic radius of

the anion )

$0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a$

Also ; a = 664 pm

Then :

$r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm$

Therefore, the ionic radius of the anion $r^- = 312.52 \ pm$

4 0

3 years ago