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3 years ago

13

Each of the following compounds has been prepared from p-nitroaniline. Outline a reasonable series of steps leading to each one.

1 answer:

Salsk061 [2.6K]3 years ago

6 0

Answer:

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Explanation:

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

4 years ago

Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal

german

Answer:

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

Explanation:

$\Delta T_b=K_b\times m$

$\Delta T_b=K_b\times \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}\times \text{Mass of chloroform in Kg}}$

where,

$\Delta T_f$ =Elevation in boiling point = $(62.5-61.7)^oC=0.8^oC$

Mass of acenapthalene = 0.515 g

Mass of $CHCl_3$ = 15.0 g = 0.015 kg (1 kg = 1000 g)

$K_b$ = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

$0.8^0C=3.67 ^oC/m\times \frac{0.515}{\text{Molar mass of acenapthalene}\times 0.015kg}$

$\text{Molar mass of acenapthalene}=155.7875 g/mol$

Let the molecule formula of the Acenapthalene be $C_{6n]H_{5n}$

$6n\times 12 g/mol+5n\times 1 g/mol=155.7875 g/mol$

n = 2.0

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

4 0

4 years ago

How to name the compounds in chemistry using roman numerals khan academy?

andrezito [222]

What are the answer choices?

7 0

3 years ago

the molar solubility of Zn(OH)2 is 5.7x 10^-3 mol/L at a certain temperature. Calculate the value of Ksp for Zn(OH)2 at this tem

Alex

Answer:

Ksp = 7.4x10⁻⁷

Explanation:

Molar solubility of a substance is defined as the amount of moles of that can be dissolved per liter of solution.

Ksp of Zn(OH)₂ is:

Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻

Ksp = [Zn²⁺] [OH⁻]²

And the molar solubility, X, is:

Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻

⇄ X + 2X

<em>Because X are moles of substance dissolved.</em>

<em>Ksp = </em>[X] [2X]²

Ksp = 4X³

As molar solubility, X, is 5.7x10⁻³mol/L:

Ksp = 4X³

Ksp = 4 (5.7x10⁻³mol/L)³

<h3>Ksp = 7.4x10⁻⁷</h3>

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3 years ago

Which process occurs at the anode in an electrochemical cell?

Fiesta28 [93]

The process that occurs at the anode in an electrochemical cell is the loss of electrons. The answer is number 2. Oxidation occurs as electrons are lost at the anode at one of the electrodes in both an electrolytic cell and a voltaic cell.

5 0

3 years ago