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3 years ago

8

Arrange the following compounds in order of increasing acidity, and explain the reasons for your choice of order. Enter your ans

wer in the answer box below. a. 4-nitrophenol b. cyclohexanol c. 2,6-difluorocyclohexanol d. 4-bromophenol

1 answer:

NNADVOKAT [17]3 years ago

6 0

Answer:

(a) > (d) > (c) > (b)

Explanation:

Acidic character is based upon inductive effect of the functional group that are attached to parent chain.

1. Nitro is a -I group which makes phenol acidic due to increased resonance, 2. bromine is a -I group but it is weaker than nitro group

3. phenol is more acidic than cyclohexanol due to resonance stabilization of resonance and among halogen F>Cl>Br>I is the order of acidity

4. cyclohexanol will be weakest as it is not associated with any -I groups here.

so the order of acidity will be

(a). 4-nitropheno > (d). 4-bromophenol > (c). 2,6-difluorocyclohexanol > (b). cyclohexanol

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An aqueous solution of hydrochloric acid is standardized by titration with a 0.137 M solution of sodium hydroxide. If 16.1 mL of

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The molarity of the hydrochloric acid, HCl solution given the data is 0.079 M

<h3>Balanced equation </h3>

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 1
The mole ratio of the base, NaOH (nB) = 1

<h3>How to determine the molarity of HCl </h3>

Molarity of base, NaOH (Mb) = 0.137 M

Volume of base, NaOH (Vb) = 16.1 mL
Volume of acid, HCl (Va) = 27.9 mL
Molarity of acid, HCl (Ma) =?

MaVa / MbVb = nA / nB

(Ma × 27.9) / (0.137 × 16.1) = 1

(Ma × 27.9) / 2.2057 = 1

Cross multiply

Ma × 27.9 = 2.2057

Divide both side by 27.9

Ma = 2.2057 / 27.9

Ma = 0.079 M

Thus, the molarity of the HCl solution is 0.079 M

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2 years ago

Vanillin, c8h8o3 (m = 152 g/mol), is the molecule responsible for the vanilla flavor in food. how many oxygen atoms are present

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Molecular weight of vanillin = 152 g/mol

Futher molecular formula of vanillin is C8H8O3

Atomic weight of oxygen = 16 g/mol

Thus, 152 g of vanillin contains 16 g of oxygen
∴ 0.045 g (45 mg) of vanillin contains $\frac{16X0.045}{152}$ = 0.00473 g

Also, number of moles of vanillin in 0.045 g sample = $\frac{weight}{molecular.weight} = \frac{0.045}{152} = 2.96X10^{-4}$
Now, 1 mole = 6.023 X 10^23 molecules
∴ 2.96 X 10^-4 mole = 1.78 X 10^20 molecules

From molecular formula, it can be seen that 1 molecule of vanallin contain 3 atoms of oxygen
∴1.78 X 10^20 molecules contain 3 X 1.78 X 10^20 = 5.34 X 10^20 oxygen atoms

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4 years ago

A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-

Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u> $Ni(s)\rightarrow Ni^{2+}+2e^-$

<u>Reduction half reaction (cathode):</u> $Ni^{2+}+2e^-\rightarrow Ni(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[Ni^{2+}_{diluted}]$ = $1.00\times 10^{-4}M$

$[Ni^{2+}_{concentrated}]$ = 1.0 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}$

$E_{cell}=0.118V$

Hence, the cell potential of the cell is +0.118 V

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3 years ago

What is the equilibrium partial pressure of water vapor above a mixture of 62.9 g H2O and 33.2 g HOCH2CH2OH at 55 °C. The partia

bija089 [108]

Answer : The partial pressure of $H_2O$ is 102.3 mmHg.

Explanation :

As per question,

Mass of $H_2O$ = 62.9 g

Mass of $HOCH_2CH_2OH$ = 33.2 g

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HOCH_2CH_2OH$ = 62 g/mole

First we have to calculate the moles of $H_2O$ and $HOCH_2CH_2OH$ .

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{62.9g}{18g/mole}=3.49mole$

$\text{Moles of }HOCH_2CH_2OH=\frac{\text{Mass of }HOCH_2CH_2OH}{\text{Molar mass of }HOCH_2CH_2OH}=\frac{33.2g}{62g/mole}=0.536mole$

Now we have to calculate the mole fraction of $H_2O$ and $HOCH_2CH_2OH$ .

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }HOCH_2CH_2OH}=\frac{3.49}{3.49+0.536}=0.867$

$\text{Mole fraction of }HOCH_2CH_2OH=\frac{\text{Moles of }HOCH_2CH_2OH}{\text{Moles of }H_2O+\text{Moles of }HOCH_2CH_2OH}=\frac{0.536}{3.49+0.536}=0.134$

Now we have to partial pressure of $H_2O$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of water vapor

$p_T$ = total pressure of gas

$X$ = mole fraction of water vapor

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.867\times 118.0mmHg=102.3mmHg$

Therefore, the partial pressure of $H_2O$ is 102.3 mmHg.

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3 years ago

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