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3 years ago

Why boiling point is more when vanderwall force if more?

Chemistry

1 answer:

jek_recluse [69]3 years ago

6 0

Answer:

It is the intermolecular forces acting between the molecules that cause attractions between them making them liquids or solids. The strength of Van der Waals forces depends primarily on the number of electrons in total in the molecule, so larger molecules will have higher boiling points.

Explanation:

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Spell out the full name of this compound:

Korvikt [17]

Answer:

nomenclature of alkanes

Explination:

Substituent Formula Number of C Atoms Name of Substituent

CH3 1 methyl–

CH3CH2 2 ethyl–

CH3CH2CH2 3 propyl–

CH3CH2CH2CH2 4 butyl–

CH3CH2CH2CH2CH2 5 pentyl–

and so forth and so forth and so forth

Try this it helped me allot,

Alkenes and alkynes are named in a similar fashion. The biggest difference is that when identifying the longest carbon chain, it must contain the C–C double or triple bond. Furthermore, when numbering the main chain, the double or triple bond gets the lowest possible number. This means that there may be longer or higher-numbered substituents than would be allowed if the molecule were an alkane. For example, this molecule is 2,4-dimethylhept-3-ene (note the number and the hyphens that indicate the position of the double bond).

2,4-dimethylhept-3-ene

or two one carbonsubstituents on the second and third C atoms

its not 3-methylheptane.

3 0

3 years ago

3.01x10^15 atoms of S= g

astra-53 [7]

There are 1.65 × 10^(-7) g S in 3.01 × 10^15 atoms S.

<em>Step 1</em>. Use Avogadro’s number to convert <em>atoms of S to moles of S</em>

Moles of S = 3.01 × 10^15 atoms S × (1 mol S/6.022 × 10^23 atoms S)

= 5.148 × 10^(-9) mol S

<em>Step 2</em>. Use the molar mass of S to convert <em>moles of S to grams of S </em>

Mass of S = 5.148 × 10^(-9) mol S × (32.06 g S/1 mol S) = 1.65 × 10^(-7) g S

4 0

4 years ago

Na and ci are chemical ( symbols,formulas).

lozanna [386]

I believe they are chemical symbols. You can find them on the periodic table.

6 0

4 years ago

Read 2 more answers

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.85×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.17×10−4, what is the

kramer

Answer:

The equilibrium constant of the given reaction is 0.01351.

Explanation:

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

The equilibrium constant of the reaction = $K_3=1.85\times 10^{-10}$

$K_3=\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}$ ...[1]

$AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$

The equilibrium constant of the reaction = $K_4=1.17\times 10^{-4}$

$K_4=\frac{[Ag^+][Cl^-]}{[AgCl]}$ ..[2]

$[Cl^-]=\frac{K_4\times [AgCl]}{[Ag^+]}$

$PbCl_2(aq)+2Ag^+(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$

The expression of equilibrium constant of the creation is ;

$K=\frac{[AgCl]^2[Pb^{2}]}{[PbCl_2]][Ag^+]^2}$

Dividing [1] by [2]

$\frac{K_3}{K_4}=\frac{\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}}{\frac{[Ag^+][Cl^-]}{[AgCl]}}$

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][Cl^-][AgCl]}{[PbCl_2][Ag^+]}$

Substituting the value of $[Cl^-]$ from [2] :

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][AgCl]}{[PbCl_2][Ag^+]}\times \frac{K_4\times [AgCl]}{[Ag^+]}$

$\frac{K_3}{K_4}=K_4\times K$

$K=\frac{K_3}{(K_4)^2}=\frac{1.85\times 10^{-10}}{(1.17\times 10^{-4})^2}$

$K=0.01351$

The equilibrium constant of the given reaction is 0.01351.

3 0

3 years ago

help asap show work | a gas with a volume of 13L at a pressure of 197 kPA is allowed to expand to a volume of 23L. what is the p

Lera25 [3.4K]

Answer:

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we're finding the new pressure

$P_2 = \frac{P_1V_1}{V_2} \\$

We have

$P_2 = \frac{13 \times 197000}{23} = \frac{2561000}{23} \\ = 111347.826...$

We have the final answer as

Hope this helps you

5 0

3 years ago