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Aleksandr-060686 [28]

3 years ago

14

Draw the product that is formed when the compound shown below is treated with an excess of hydrogen gas and a platinum catalyst.

hexene

1 answer:

loris [4]3 years ago

5 0

<span>When the hexene is treated with an excess of hydrogen gas and a platinum catalyst, the product formed is hexane.

Hexene is an unsaturated hydrocarbon. Hexene when treated with H2 gas, in presence with Pt catalyst, the double bond is broken. This results in generation of saturated hydrocarbon i.e. hexane. </span>

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Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002

marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The $pK_a=4.756$

The value of the dissociation constant = $K_a$

$pK_a=-\log[K_a]$

$K_a=10^{-4.756}=1.754\times 10^{-5}$

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

$HAc\rightleftharpoons H^++Ac^-$

Initially

c

At equilibrium ;

(c-cα) cα cα

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][Ac^-]}{[HAc]}$

$1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}$

$1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}$

$1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}$

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

$[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M$

The pH of the solution ;

$pH=-\log[H^+]$

$=-\log[0.0001901 M]=3.72$

3 0

3 years ago

PLEASE HELP ASAP

erma4kov [3.2K]

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6 0

3 years ago

How does a sciences form a hypothesis

Oduvanchick [21]

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7 0

2 years ago

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Americans combined drive about 4.0 x 109 kilometers a day and get an average of 20 miles per gallon of gasoline. For each kilogr

Nimfa-mama [501]

Answer:

$303,882.84649 kg\times 3=9.12\times 10^5 kg$ of carbon dioxide gas.

Explanation:

Average distance covered by Americans in a day= $4.0\times 10^9 km$

1 day = 24 × 60 min = 1,440 min

Average distance covered by Americans in a minute= $\frac{4.0\times 10^9 km}{1,440}=2,777,777.78 km$

Average mileage of the car = 20 miles/gal = 32.18 km/gal

1 mile = 1.609 km

20 miles = 20 × 1.609 km = 32.18 km

Volume of gasoline used in minute = $\frac{2,777,777.78 km}{32.18 km/gal}$

$V=86,320.00 gal$

$V=86,320.00\times 3.7854 L$

(1 L = 1000 mL)

$V=86,320.00\times 3.7854 \times 1000 mL=326,755,748.91 mL$

Mass of 86,320.00 gallons of gasoline = m

Density of the gasoline = d = $0.93 g/cm^3=0.93 g/mL$

$1 mL= 1 cm^3$

$m=d\times V=0.93 g/mL\times 326,755,748.91 mL$

$m=303,882,846.49 g=303,882.84649 kg$

1 kilogram of gasoline gives 3 kg of carbon dioxde gas .

Then 303,882.84649 kg of gasoline will give :

$303,882.84649 kg\times 3=9.12\times 10^5 kg$ of carbon dioxide gas.

8 0

3 years ago

109kPa = _________ mm Hg

LekaFEV [45]

817.567 mm hg the answer for number 2

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3 years ago