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balu736 [363]
3 years ago
13

Question 4

Chemistry
1 answer:
ikadub [295]3 years ago
4 0

Answer:,

Explanation:,

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The specific gravity of water is 1.0. True False
Zielflug [23.3K]
The specific gravity of water is 1.0 True 
5 0
3 years ago
Compare and contrast the molecular activity of solids, liquids, and gases.​
cupoosta [38]

Answer:

The molecules in a solid are close together and they vibrate in place. The molecules in a liquid move quickly and farther apart from eachother. The particles in a gas move freely at high speeds and slide past eachother.

5 0
3 years ago
Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi
sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Entropy change of the reaction ; ΔS

\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]

=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

6 0
3 years ago
A solid sample contains both NaOH and NaCl. 0.500 g of this solid sample was dissolved in water to make a 20.0 mL solution and t
Setler [38]

Explanation:

Below is an attachment containing the solution.

3 0
4 years ago
Nitrogen gas and chlorine gas will react to form nitrogen monochloride gas. 5 moles of nitrogen and 10 moles of chlorine are mix
raketka [301]

Answer:

0.30

Explanation:

Step 1: Calculate the known initial and equilibrium molar concentrations

[N₂]i = 5 mol/2 L = 2.5 M

[Cl₂]i = 10 mol/2 L = 5 M

[NCl]e = 3 mol/2 L = 1.5 M

Step 2: Make an ICE chart

        N₂(g) + Cl₂(g) ⇄ 2 NCl(g)

I         2.5       5                0

C        -x        -x              +2x

E      2.5-x    5-x               2x

Step 3: Find the value of x

We know that [NCl]e = 2x = 1.5 ⇒ x = 0.75 M

Step 4: Calculate the concentrations at equilibrium

[N₂]e = 2.5-x = 2.5-0.75 = 1.75 M

[Cl₂]e = 5-x = 5-0.75 = 4.25 M

[NCl]e = 1.5 M

Step 5: Calculate the equilibrium constant (Kc)

Kc = [NCl]² / [N₂] × [Cl₂]

Kc = 1.5² / 1.75 × 4.25 = 0.30

7 0
3 years ago
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