All categories

3 years ago

Doing which of the following generally increases entropy of a substance

1 answer:

8 0

Freezing the substance and condensing it will not increase the entropy of a substance. In fact, dissolving the substance into water is the only way to increase the entropy of a certain substance. The entropy increases because it diversify with each other.

You might be interested in

If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?

Fiesta28 [93]

The given sentence is part of a longer question.

I found this question with the same sentence. So, I will help you using this question:

For the reaction N2O4(g) ⇄ 2NO2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm). If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?

(a) Q = K; The reaction is at equilibrium.
(b) Q < K; The reaction will proceed to the right.
(c) Q > K; The reaction will proceed to the left.

The answer is the option (c) Q > K; The reaction will proceed to the left, since Qp = 1, and 1 > 0.15.
Explanation:

Kp is the equilibrium constant in term of the partial pressures of the gases.

Q is the reaction quotient. It is a measure of the progress of a chemical reaction.

The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.

At equilibrium both Kp and Q are equal. Q = Kp

If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,

If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.

Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.

Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.

6 0

3 years ago

The body gets rid of lactic acid in a chemical pathway that requires ___________. A. carbon dioxide . B. oxygen . C. amino acids

Luden [163]

The body gets rid of acid in a chemical pathway that requires oxygen. The correct answer is B, oxygen.

5 0

3 years ago

Please answer both questions.

Deffense [45]

Answer:

1)Krypton

2)11H

Explanation:

electrons=protons

protons=atomic number

mass number=protons+neutrons

mass number is the superscript

atomic number is the subscript.

1)The answer is Krypton because its atomic number= number of protons=number of electrons is 36.

mass number is 46+36=82.

2)subscript=atomic number=number of protons=number of electrons

i. H = electrons=1

=neutrons=0

ii. Cl=electrons=17

=neutrons=35-17=18

iii. Na=electrons=11

=neutrons=23-11= 12

so the answer is Hydrogen because it has 1 electron and 0 neutron.

I hope this helps.

6 0

3 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

3 years ago

What tuberculosis means "

labwork [276]

Answer:

it is a infectiousr bacterial disease characterized by the growth of nodules(tubercles) in tissues especially the lungs

3 0

3 years ago