Gather data: experiment with a variety of initial concentrations of no2 and n2o4. for each set of initial concentrations, use the gizmo to determine the equilibrium concentrations of each substance. in the last column, find kc for that trial. run three trials for each set of initial conditions.

Answers

Answer 1

Solution :

[tex]$H_2 +I_2\rightleftharpoons 2HI$[/tex]

Initial concentration               Equilibrium concentration           [tex]$K_c=\frac{[HI]^2}{[H_2][I_2]}$[/tex]

  (mol/L)                                           (mol/L)

[tex]$[H_2]$[/tex]                 [tex]$[I_2]$[/tex]            [tex]$[HI]$[/tex]    [tex]$[H_2]$[/tex]                 [tex]$[I_2]$[/tex]            [tex]$[HI]$[/tex]

[tex]$2.4 \times 10^{-2}$[/tex]     [tex]$1.38 \times 10^{-2}$[/tex]  0     [tex]$1.14 \times10^{-2}$[/tex]    [tex]$0.12 \times 10^{-2}$[/tex]  [tex]$2.52 \times 10^{-2}$[/tex]        46.42

[tex]$2.4 \times 10^{-2}$[/tex]     [tex]$1.68 \times 10^{-2}$[/tex]  0     [tex]$0.92 \times10^{-2}$[/tex]    [tex]$0.20 \times 10^{-2}$[/tex]  [tex]$2.96 \times 10^{-2}$[/tex]        47.61

[tex]$2.4 \times 10^{-2}$[/tex]     [tex]$1.98 \times 10^{-2}$[/tex]  0     [tex]$0.77 \times10^{-2}$[/tex]    [tex]$0.31 \times 10^{-2}$[/tex]  [tex]$3.34 \times 10^{-2}$[/tex]        46.73

[tex]$2.4 \times 10^{-2}$[/tex]     [tex]$1.76 \times 10^{-2}$[/tex]  0     [tex]$0.92 \times10^{-2}$[/tex]    [tex]$0.22 \times 10^{-2}$[/tex]  [tex]$3.08 \times 10^{-2}$[/tex]        46.86

[tex]$0$[/tex]                  [tex]$0$[/tex]        [tex]$3.04 \times 10^{-2}$[/tex]     [tex]$0.345 \times10^{-2}$[/tex]    [tex]$0.345 \times 10^{-2}$[/tex]  [tex]$2.35 \times 10^{-2}$[/tex] 46.39

[tex]$0$[/tex]                  [tex]$0$[/tex]        [tex]$7.58 \times 10^{-2}$[/tex]     [tex]$0.86 \times10^{-2}$[/tex]    [tex]$0.86 \times 10^{-2}$[/tex]  [tex]$5.86 \times 10^{-2}$[/tex]     46.42

Average [tex]$K_c=46.738$[/tex]


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-wikipedia

Explanation:

[tex]\huge{\textbf{\textsf{{\color{pink}{An}}{\red{sw}}{\orange{er}} {\color{yellow}{:}}}}}[/tex]

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Answer:

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Answers

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Answers

Answer:

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Answers

P2 = 54.6 kPa

Explanation:

Given:

V1 = 10.0 L. V2 = 50.0 L

P1 = 273 kPa. P2 = ?

We can use Boyle's law to solve this problem.

P1V1 = P2V2

Solving for P2,

P2 = (V1/V2)P1

= (10.0 L/50.0 L)(273 kPa)

= 54.6 kPa

Look at the diagram below. According to the diagram, what substance(s) are the
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Answers

Answer:

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Explanation:

The hydrogen atom from the HCl molecule joins the other three in NH3 creating a four hydrogen molecule,  NH4Cl.

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Answer:

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We have the final answer as

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Answers

Answer Explanation:

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Answers

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Answer:

All of these answers are correct.

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Explanation:

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A gas diffuses 1/7 times faster than hydrogen gas (H2).

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Answers

Answer: The molar mass of the gas is 9.878 g/mol.

Explanation:

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M = molar mass of gas

As given gas diffuses 1/7 times faster than hydrogen gas. So, its molar mass is calculated as follows.

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where,

[tex]M_{1}[/tex] = molar mass of hydrogen gas

[tex]M_{2}[/tex] = molar mass of another given gas

[tex]R_{1}[/tex] = rate of diffusion of hydrogen

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Substitute the values into above formula as follows.

[tex]\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\\\frac{R_{1}}{\frac{1}{7}R_{1}} = \sqrt{\frac{M_{2}}{2}}\\7 \times 1.414 = M_{2}\\M_{2} = 9.878 g/mol[/tex]

Thus, we can conclude that the molar mass of the gas is 9.878 g/mol.

Answer:

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Explanation:

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Answer:

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0.199 mol KMnO4
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Answers

The answer is: 0.158 mol
You find this by doing:
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n=158.034/25.0

Answer:

158 mol :)

Explanation:

How many calories are absorbed in a process that absorbs 0.128 joules?
A. 0.031 cal
B. 0.536 cal
C. 536 cal
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Answers

The answer is a - round up 0.0305 to 0.031

Atoms that have become negatively charged by gaining extra electrons are called

Answers

Answer:

CATION

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Given the following equation: Na2O + H2O ---> 2 NaOH
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Answers

Answer:

2.04 moles

Explanation:

Since the ratio of Na2O to NaOH is 1:2, you multiply the number of moles by 2, equaling 2.04 moles.

Imagine you had HCl with a concentration of exactly 0.10 mol/dm3. If 0.023 dm3 of a sodium hydroxide solution, NaOH (aq), could exactly neutralize 0.040 dm3 of the HCl solution, what is the concentration of the NaOH (aq)

Answers

Answer:

Explanation:

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3

Rearrange:

Concentration in mol/dm3 =

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3

Concentration in mol/dm3 =

Concentration in mol/dm3 =

= 0.125 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.125

= 4.56 g

So concentration = 4.56 g/dm3

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