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3 years ago

Draw the structure of 2 methyl,but-3-ene

Chemistry

1 answer:

Maslowich3 years ago

5 0

Answer:

Please find the structure attached as an image

Explanation:

Based on the characteristics ending name (-ene) of the organic compound above, it belongs to the ALKENE GROUP. Alkenes are characterized by the possession of a carbon to carbon double bond (C=C) in their structure.

- But-3-ene tells us that the organic compound has four straight carbon atoms with the C=C (double bond) located on the THIRD carbon depending on if we count from right to left or vice versa.

- 2 methyl indicates that the methyl group (-CH3) is located as an attachment on the second carbon (carbon 2).

N.B: In the structure attached below, the counting is from the left to right (→).

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Answer:

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3 years ago

In a coffee cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00 degrees C. After

igomit [66]

Answer:

+26.6kJ/mol

Explanation:

The enthalpy of dissolution of NH₄NO₃ is:

NH₄NO₃(aq) + ΔH → NH₄⁺ + NO₃⁻

Where ΔH is the heat of reaction that is absorbed per mole of NH₄NO₃,

The moles that reacts in 1.60g are (Molar mass NH₄NO₃:80g/mol):

1.60g * * (1mol / 80g) = 0.02 moles reacts

To find the heat released in the coffee cup calorimeter, we must use the equation:

Q = m×ΔT×C

Where Q is heat released,

m is mass of the solution

ΔT is change in temperature (Final temperature - Initial temperature)

C is specific heat of the solution (4.18J/g°C)

Mass of the solution is:

1.60g + 75g = 76.60g

Change in temperature is:

25.00°C - 23.34°C = 1.66°C

Replacing:

Q = m×ΔT×C

Q = 76.60g×1.66°C×4.18J/g°C

Q = 531.5J

This is the heat released per 0.02mol. The heat released per mole (Enthalpy change for the dissolution of NH₄NO₃) is:

531.5J / 0.02mol = 26576J/ mol =

+26.6kJ/mol

<em>+ because the heat is absorbed, the reaction is endothermic-</em>

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3 years ago

What causes the shielding effect to remain constant across a period?

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Electrons are added to the same principal energy level.

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4 years ago

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

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