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2 years ago

8

Which molecule below has hydrogen bonding? Which molecule below has hydrogen bonding? H2 CH3CH2OH HCl CH4 all of the above

1 answer:

QveST [7]2 years ago

3 0

Answer:

$CH_3CH_2OH$

Explanation:

Hydrogen bonding:-

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Partially positive end of the hydrogen atom is attracted to partially negative end of these atoms which is present in another molecule. It is strong force of attraction between the molecules.

Thus, hydrogen must be linked to electronegative atom which is oxygen, fluorine and nitrogen which is in $CH_3CH_2OH$ and thus, it will shown hydrogen bonding.

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4Al(s)+3O2(g)__> 2Al2O3(s)

nasty-shy [4]

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3 years ago

Which statement summarizes the difference between mass and weight?

alexgriva [62]

Answer:

Mass is the amount of matter in an object.

Weight is how much an object weighs.

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3 years ago

A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

2 years ago

Read 2 more answers

At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1·M−1s−1 : →2SO3g+2SO2gO2g Suppo

RoseWind [281]

Answer:

$[SO_3]=0.25M$

Explanation:

Hello there!

In this case, since the integrated rate law for a second-order reaction is:

$[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}$

Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:

$[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\$

$[SO_3]=0.25M$

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4 0

2 years ago

Calculate the missing variables in each experiment below using Avogadro’s law.

blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml n1 = 9.11 mol

V2 = 123 ml n2 = ?

Formula

$\frac{V1}{n1} = \frac{V2}{n2}$

$n2 = \frac{n1V2}{V1}$

$n2 = \frac{(9.11)((123)}{(612)}$

n2 = 1.83 mol

5 0

2 years ago