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svetlana [45]
2 years ago
10

or the solution containing Zn2+(aq), n is 2, so the MM/n ratio will be the molar mass of Zn divided by 2: MM/n = (65.39 g/mol) /

2 = 32.70 g/mol Compare this theoretical value with your experimental result. If the plating efficiency is significantly less than 1, say it is 0.55, how would the observed MM/n ratio compare to the theoretical value?
Chemistry
1 answer:
yKpoI14uk [10]2 years ago
4 0

Answer:

MM/n will be less than the theoretical value

Explanation:

Provided that

The plating efficiency is proportional to the element weight

i.e.

\frac{Zn^2}{(aq)}

That represents \frac{MM}{n\ value}

So,

MM by N value is

= 65.39 g /mol ÷ 2

= 32.70 g/mol

Also, the experimental plating value is 0.55 that represents that it is much lesser than the normal value

So it would be concluded that the MM by N would be lower than the theoretical value

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Water is an example of a compound. Which of the following is the best description of all compounds?
AlexFokin [52]

Answer:

A compound has atoms of different elements chemically joined together They can't be separated without a chemical reaction.

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identify the reagents necessary to achieve each of the following transformations o3 dms, pcc, ch2cl, h2so4,h2o,hgso4
hoa [83]

the reagents necessary to convert alcohol to ketoneNa_2Cr_2O_7 , H_2O which involves oxidation of alcohols.

<h3>What is oxidation of alcohols?</h3>
  • Alcohol oxidation is a significant organic chemistry process. Secondary alcohols can be oxidized to produce ketones, while primary alcohols can be oxidized to produce aldehydes and carboxylic acids.
  • In contrast, tertiary alcohols cannot be oxidized without the C-C bonds in the molecule being broken.
  • In order to cause primary alcohols to oxidize into aldehydes
  1. Cr_2O_7 ^-^2 (dichromate)
  2. CrO_3/pyridine (Collins reagent)
  3. Chromium pyridinium compound (PCC)
  4. Dichromate of pyridinium (PDC, Cornforth reagent)
  5. Periodinane by Dess-Martin
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  • oxidation of secondary alcohols to ketones
  1. Cr_2O_7 ^-^2 (dichromate)
  2. CrO_3/pyridine (Collins reagent)
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To learn more about oxidation of alcohols with the given link

brainly.com/question/7207863

#SPJ4

<u>Question:</u>

Identify the reagents necessary to achieve each of the following transformations

a. O_3 , DMS

b.H_2SO_4 , H_2O , HgSO_4

c.Na_2Cr_2O_7 , H_2O

d.Fe ^ {2+}, NaOH

3 0
2 years ago
A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513
igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is:  N₂O₄(g)  ⇆  2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

              2NO₂ (g)   ⇆   N₂O₄(g)      

Initially   2.20 mol              -

React          x                      x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq       (2.20-x) / 3.50L     (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) /  [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0  → Quadratic formula

a = 0.2933 ;  b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50  = 0.0971 M

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Licemer1 [7]

Answer:

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Explanation:

The number of moles of solute is equal to product of the molar concentration (molarity) and the volume (in liters) of solution.

Since the volumes and the molar concentrations of the<em> NaOH </em>and <em>HCl </em>solutions mixed are equal, each one of them contributes the same number of moles of solute.

Since every mol of NaOH produces one mol of OH⁻ ions and every mol of HCl produces one mol of H⁺ ion, the number of moles of OH ⁻ and H⁺ in solution are equal.

Thus, OH⁻ and H⁺ ions will be neutralized by the reaction:

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Which is strongly shifted to the right and has <em>neutral pH</em>.

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6 0
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