All categories

3 years ago

2. Which pair of elements does not form an ionic bond?

Chemistry

1 answer:

Gekata [30.6K]3 years ago

5 0

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Bond Formation.

Answer is :-

B.) O and Cl

These both are Non-Metals with high Electronegativity nature, hence none of them can share electrons to foram any bond to eventually form a Electro valentines Compound.

You might be interested in

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

Degger [83]

a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3

8 0

3 years ago

Read 2 more answers

What would the formula of gallium bromide be?

Free_Kalibri [48]

GaBr3

Gallium=Ga
Bromine= Br
Bromide=Br3

5 0

3 years ago

The equilibrium concentrations of the reactants and products are [ HA ] = 0.260 M [HA]=0.260 M , [ H + ] = 2.00 × 10 − 4 M [H+]=

ValentinkaMS [17]

Answer:

pKa of the acid HA with given equilibrium concentrations is 6.8

Explanation:

The dissolution reaction is:

HA ⇔ H⁺ + A⁻

So at equilibrium, Ka is calculated as below

Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260

= 15.38 x 10⁻⁸

Hence, by definition,

pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813

7 0

3 years ago

Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.

goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

$P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}$

In the first solution:

$X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625$

$X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375$

$340torr = 0.625P^0_{A}+0.375P^0_{B}$ (1)

For the second equation:

$X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700$

$X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300$

$370torr = 0.700P^0_{A}+0.300P^0_{B}$ (2)

Replacing (2) in (1):

$340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})$

$340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}$

-122.5torr = -0.250P°A

$P^0_{A} = 490 torr$

Vapour pressure of cyclohexane at 50°C is 490torr

And for benzene:

$370torr = 0.700*490torr+0.300P^0_{B}$

$P^0_{B}=90torr$

Vapour pressure of benzene at 50°C is 90torr

3 0

3 years ago

Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate

Andrew [12]

Answer:

(2R,3S)-2-ethoxy-3-methylpentane

and

(2S,3S)-2-ethoxy-3-methylpentane

Explanation:

For this case, we will have $CH_3CH_2O^-$ as nucleophile. Also, this compound is also in excess. So, we will have as solvent $CH_3CH_2OH$ a protic solvent. Therefore the Sn1 reaction would be favored.

The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).

7 0

3 years ago