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3 years ago

What is the basic of an organic molecule

Chemistry

2 answers:

julia-pushkina [17]3 years ago

7 0

proteins, carbohydrates, lipids and nucleic acids.

Deffense [45]3 years ago

4 0

Organic molecules are the molecules of life and are built around chains of carbon atoms that are often quite long. There are four main groups of organic moleculesthat combine to build cells and their parts: carbohydrates, proteins, lipids, and nucleic acids.What is the basic of an organic molecule

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Given the equilibrium constants for the following two reactions at a 298K:NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40NiO(s) +CO(g) ⇌ N

shusha [124]

Answer:

The value is $K_C = \frac{40}{600}$

Explanation:

From the question

The equation given is

NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40 (1)

NiO(s) +CO(g) ⇌ Ni(s) +CO2(g) Kc=600 (2)

Generally the reverse of the second equation as shown below

Ni(s) +CO2(g) ⇌ NiO(s) + CO(g) (3)

The equilibrium constant becomes $K_c ' = \frac{1}{600}$

Now adding 1 and 3 we obtain

NiO(s) + H2(g)+ Ni(s) +CO2(g) ⇌ Ni(s) + H2O(g)+NiO(s) + CO(g)

CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

Hence the equilibrium constant for the resulting equation is mathematically evaluated as

$K_C = K_c' * K_c$

=> $K_C = \frac{1}{600} * 40$

=> $K_C = \frac{40}{600}$

6 0

3 years ago

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$

Taking inverse log both side we get,

$\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg$

8 0

3 years ago

Which of the following results in an increase in the entropy? (4 points)

Dima020 [189]

Dissolving sugar in water results in an increase in the entropy.

Hence, Option (D) is correct answer.

<h3>What is Entropy ?</h3>

Measurement of randomness of a system is called entropy. It is an extensive property. It is a state function. Unit of entropy is JK⁻¹ mol⁻¹.

Now lets check all options one by one

Option (A): Freezing water

Freezing water decreases the entropy because here second law of thermodynamics does not violate.

So it is incorrect option

Option (B): Cooling water

Cooling water does not increases entropy because entropy increases when solid melts to give liquid.

So it is incorrect option

Option (C): Condensing water vapour

In Condensing water vapour the temperature of liquid phase decreases and thus kinetic energy decreases. The randomness will decrease and hence entropy will also decrease.

So it is incorrect option.

Option (D): Dissolving sugar in solute

In dissolving sugar in solute the solid dissociates to ions and the randomness will increase and hence entropy will also increase.

So it is correct option

Thus from the above conclusion we can say that Dissolving sugar in water results in an increase in the entropy.

Hence, Option (D) is correct answer.

Learn more about the Entropy here: brainly.com/question/1477087

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2 years ago

Question 16 of 33

Shtirlitz [24]

I think the answer is A

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3 years ago

Which of the following scenarios is representative of parasitism?

Alex73 [517]

What are the following answers?

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4 years ago

Read 2 more answers