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3 years ago

Which of the following represents a hydrocarbon that contains one double bond somewhere in the carbon chain?

1 answer:

8 0

The answer is alkene.

An alkane is a saturated hydrocarbon, thi is it has only single bonds.

Alkenes and alkynes are unsaturated: alkenes have double bonds and alkynes have triple bonds.

Subsituted hydrocarbon, is a hydrocarbon with one hydrogen substituted by another element or a group.

For example:

CH3 - CH - CH3
|
F

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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

Can somebody help me with this also plz

just olya [345]

ANSWER:

4 a) Specific elements have more than one oxidation state, demonstrating variable valency.

For example, the following transition metals demonstrate varied valence states: $Fe^{2+}$ , $Fe^{3+}$ , $Cr^{2+}, Cr^{3+}$ , etc.

Normal metals such as $Pb^{2+} and Pb^{4+}$ also show variable valencies. Certain non-metals are also found to show more than one valence state $Pb^{3+} and Pb^{5+}.$

4 b) Isotopes are members of a family of an element that all have the same number of protons but different numbers of neutrons.

For example, Carbon-14 is a naturally occurring radioactive isotope of carbon, having six protons and eight neutrons in the nucleus. However, C-14 does not last forever and there will come a time when it loses its extra neutrons and becomes Carbon-12.

5 a) $2Fe + 3Cl_2$ → $2FeCl_3$

5 b) $3Pb + 8HNO_3$ → $3Pb (NO_3)_2 + 4H_2O + 2NO_2$

5 c) $Zn + H_2SO_4$ → $ZnSO_4 + H_2$ (already balanced so don't need to change)

5 d) $2H_2 + O_2$ → $2H_2O$

5 e) $2Mg + 2HCl$ → $2MgCl + H_2$

EXPLANATION (IF NEEDED):

1. Write out how many atoms of each element is on the left (reactant side) and right (product side) of the arrow.

2. Start multiplying each side accordingly to try to get atoms of the elements on both sides equal.

EXAMPLE OF BALANCING:

8 0

2 years ago

The graph shows the number of beans eaten by a random zocco. How many light red kidney beans were eaten on Day 7? How many dark

Tems11 [23]

Answer:

day 7 50 and day 8 40

Explanation:

3 0

3 years ago

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What is the most common substance found in natural gas

kifflom [539]

Methane composes most of the natural gas.

8 0

3 years ago

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Describe an acid or base in terms of its relationship to protons

motikmotik

Acids are donors of protons (H+) and bases are acceptors of protons.
For example:
1) hydrochloric acid (HCl) in reaction with water give one proton to water and become chloride anion (Cl-).
2) ammonia (NH3) Is base, in reaction with water accepts one protone and become ammonium cation (NH4+).

4 0

3 years ago