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Savatey [412]
2 years ago
10

The value of the entropy change for the process N₂ (g) + 3H₂ (g) --> 2NH₃ (g) is ________.

Chemistry
1 answer:
kodGreya [7K]2 years ago
5 0

Answer:

negative

Explanation:

Entropy is a measure of the "disorder" in a system.

In this reaction, the amount of disorder decreases. This is because one gas molecule (NH₃) has more order than two gas molecules (N₂ and H₂). Therefore, the entropy change should be negative.

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2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na2C2O¬4, is the sodium salt of this acid.
AnnZ [28]

Answer:

Sodium oxalate is a basic salt. In water it can be dissolved and dissociated.

The oxalic acid in water has two dissociations.

Explanation:

Na2C2O4 ---> 2Na+   +  C2O4-2

Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.

C2O4-2  + H2O ⇄  HC2O4-  +  OH-

In water the salt is basic.  The pH of an aqueous solution of this salt is basic, since OH- is generated.

The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.

HC2O4-  +  H2O ⇄  H2C2O4  +  OH-

The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).

H2C2O4  + H2O ⇄ H3O+  + HC2O4-

HC2O4-  +  H2O ⇄  H3O+  C2O4-2

The  HC2O4-  acts as an ampholyte since it accepts and delivers protons simultaneously.

6 0
3 years ago
A sample of gas of 752 torr and occupies a volume of 5.12 L. What will the new volume be when the pressure is increased to 1.5 A
Katyanochek1 [597]
<span> 7.80 hop it help u ok</span>
3 0
4 years ago
Can someone help me? It needs to have a diagram that has arrows.
daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)

The expression for enthalpy change is,

\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]

Putting the values we get :

\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]

-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]

H_f_{C_4H_{10}=-125kJ/mol

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0
3 years ago
(a) What structural feature is associated with each type of hydrocarbon: alkane, cycloalkane, alkene, and alkyne?
Snowcat [4.5K]

Alkanes are hydrocarbons with straight, saturated branch chains. Ring-shaped hydrocarbons are cycloalkanes. Alkenes are branch chains that are straight and have at least one double bond. Alkynes are branch chains that are straight and have at least one triple bond.

<h3>What is Hydrocarbon ?</h3>

A hydrocarbon is an organic molecule in organic chemistry that is made completely of hydrogen and carbon. Examples of group 14 hydrides include hydrocarbons. The majority of hydrocarbons are colorless and hydrophobic, and their scents are either insignificant or best characterized by those of gasoline and lighter fluid.

Other side effects from certain hydrocarbons include coma, seizures, abnormal cardiac rhythms, and liver or kidney damage. Some solvents used in paints, dry cleaning, and household cleaning solutions are examples of items that contain hazardous hydrocarbons.

To know more about Hydrocarbon please click here : brainly.com/question/3551546

#SPJ4

8 0
1 year ago
In lab (write this down in your lab protocol), you will be given a stock solution that has a glucose concentration of 60 mg/dL.
Wittaler [7]

Answer:

1. The dilution factor for the serial dilution = 2

2. V2 = 1 mL

3. V1 = 0.5 mL

Explanation:

1. Dilution factor is the ratio of the initial concentration to the final concentration.

Dilution factor = initial concentration / final concentration

First dilution: initial concentration = 60 mg/dL

final concentration = 30 mg/dL

Dilution factor = 60 mg/dL / 30 mg/dL = 2

Second dilution: initial concentration = 30 mg/dL

final concentration = 15 mg/dL

Dilution factor = 30 mg/dL / 15 mg/dL = 2

Therefore, the dilution factor for the serial dilution = 2

2. From the dilution formula, C1V1 = C2V2; V2 = final volume to be prepared.

Since 1 mL of the various glucose solutions are to be prepared, the final concentration, V2 = 1 mL

3. From the dilution formula, C1V1 = C2V2; V1 = initial concentration of the solution to be prepared.

C1/C2 = V2/V1

Since the dilution factor, C1/C2 is 2, V2/V1 = 2

V1 = V2/2

V1 = 1 mL / 2

V1 = 0.5 mL

6 0
3 years ago
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