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3 years ago

How did organic compounds get their name? Explain how the word is related to its meaning.

1 answer:

4 0

The correct answer is that organic compounds get their name from its association with living organisms. The word "organic" means living. That's how they are related and that's how organic compounds get their name. I hope this answer helped you.

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The heat of a reaction may be found with the equation q=mcΔT. A hot pot is set on a 2.51 kg slab of granite. The temperature of

bagirrra123 [75]

I believe the answer to that is c!!!

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3 years ago

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Fe2O3 + 2Al = 2Fe + Al2O3 is this a redox reaction

katovenus [111]

Answer:

<h3>2Al+ Fe2O3 gives 2Fe + Al2O3. The given reaction is a redox reaction. As oxidation and reduction are taking place simultaneously.</h3>

Explanation:

like this...Identify oxidation and reduction with their agents:

<h3>•Fe2O3 is reduced to Fe whereas Al is oxidized to Al2O3</h3>

<h3>In the above reaction:</h3>

<h3>Oxidizing agent:Fe2O3</h3>

<h3>Reducing agent:Al</h3>

I hope it's help you (◠‿・)—☆

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2 years ago

What are some of the activities performed during experiment

aev [14]

Answer:

nobmelonisegxfixcyctGkchkcigdtidtifyoc

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2 years ago

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

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2 years ago

Hamood backwards is hamood

ad-work [718]

Answer:

ano po yung sinasabi nyo

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2 years ago

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