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3 years ago

Which of the following classes of organic compounds has C=C in it?

1 answer:

4 0

The class of organic compounds that contain carbon-carbon double bonds, represented as C=C, is known as A) alkene.
Alkenes are unsaturated organic compounds and contain two carbon atoms bonded by one sigma bond and one pi bond. Alkenes exhibit different properties from alkanes due to the negative charge concentration at the double bond. Alkanes may be converted to alkenes and vice-versa.

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You have prepared a saturated solution of x at 20∘c using 39.0 g of water. how much more solute can be dissolved if the temperat

AleksAgata [21]

for 39g water solute dissolved at 20C = solubility ( g/ 100 g H2O ) × mass of water = ( 11g / 100g H2O ) × 39g H2O = 4.29 g

amount of solute dissolved at 30 C =

= 23 / 100 * 39 = 8.97 g

Amount of extra solute dissolved = 8.97 - 4.29 = 4.7 g

3 0

3 years ago

Ascorbic acid (vitamin

svlad2 [7]

Let us see the structure of ascorbic acid

As shown there is no COOH group however the OH group can lose a proton and forms conjugate base

The conjugate base formed is stabilized due to resonance

More the stability of conjugate base more the strength of acid

Hence ascorbic acid behaves as an acid

8 0

3 years ago

A sample contains 25% parent isotope and 75% daughter isotopes. If the half-life of the parent isotope is 72 years, how old is t

alisha [4.7K]

The radioactive decay obeys first order kinetics

the rate law expression for radioactive decay is

$ln\frac{[A_{0}]}{[A_{t}]}=kt$

Where

A0 = initial concentration

At = concentration after time "t"

t = time

k = rate constant

For first order reaction the relation between rate constant and half life is:

$k=\frac{0.693}{t_{\frac{1}{2} } }$

Let us calculate k

k = 0.693 / 72 = 0.009625 years⁻¹

Given

At = 0.25 A0

$ln(\frac{A0}{0.25A0})=0.009625 X time$

time = 144 years

So after 144 years the sample contains 25% parent isotope and 75% daughter isotopes**

Simply two half lives

5 0

2 years ago

Read 2 more answers

WHY IS THE OCEAN SALTY ALSO I CANT TAKE OF CAP LOCKS ON A CROME BOOK

serious [3.7K]

Answer:

What happens is,that,when a river flows into the ocean it carries some land with it. This land is made up of rocks and soil, which contains minerals. One of these minerals is Salt.

Explanation:

6 0

3 years ago

A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc

Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

7 0

2 years ago