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sveticcg [70]
4 years ago
6

The equilibrium constant has been estimated to be 0.12 at 25 °C. If you had originally placed 0.069 mol of cyclohexane in a 2.8

L flask, what would be the concentrations of cyclohexane and methylcyclopentane when equilibrium is established?
Chemistry
1 answer:
scZoUnD [109]4 years ago
4 0

Answer: Concentrations of cyclohexane and methylcyclopentane at equilibrium are 0.0223 M and 0.0027 M respectively

Explanation:

Moles of cyclohexane = 0.069 mole

Volume of solution = 2.8 L

Initial concentration of cyclohexane =\frac{moles}{Volume}=\frac{0.069}{2.8}=0.025M

The given balanced equilibrium reaction is,

                            cyclohexane  ⇔  methylcyclopentane

Initial conc.                 0.025 M           0

At eqm. conc.       (0.025-x)M       (x) M

The expression for equilibrium constant for this reaction will be,

K= methylcyclopentane / cyclohexane

Now put all the given values in this expression, we get :

0.12=\frac{(x)}{(0.025-x)}

By solving the term 'x', we get :

x =  0.0027

Concentration of cyclohexane at equilibrium = (0.025-x ) M = (0.025-0.0027) M = 0.0223 M

Concentration of methylcyclopentane at equilibrium = (x ) M = (0.0027) M

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ss7ja [257]

Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

8 0
3 years ago
Which isotope has the greatest number of protons?
tino4ka555 [31]

Answer:

The isotope with the greatest number of protons is:

  • <u>option D:  Pu-239, with 94 protons</u>

Explanation:

The number of <em>protons</em> is the atomic number and is a unique number for each type of element.

You can tell the number of protons searching the element in a periodic table and reading its atomic number.

Thus, this is how you tell the number of protons or each isotope

Sample       Chemical symbol  Element       atomic number   # of protons

A Pa-238       Pa                         protactinium         91                        91

B U-240         U                          uranium                 92                       92

C Np-238       Np                        neptunium            93                       93

D Pu-239        Pu                        plutonium              94                       94

8 0
3 years ago
What are particles that are close together and locked in place??
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Particles that are close together and locked in a place means its
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3 0
3 years ago
Draw the structure of the organic product(s) of the grignard reaction between phosgene (clcocl) and excess phenylmagnesium bromi
Jobisdone [24]
Phosgene on reacting with <span>phenylmagnesium bromide generates benzoyl chloride. 

Since, </span>phenylmagnesium bromide is added in excess. It would further react with benzoyl chloride to form benzophenone.

Benzophenone on further reacting with phenylmagnesium bromide, and aqueous treatment, gives triphenylmethanol. 

Entire reaction pathways is shown below:

8 0
3 years ago
Read 2 more answers
Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of
jeka57 [31]

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = 2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x

2330.35x=1048815\\x=450.068g

Hence, there was 450.068g of water in the pot.

8 0
3 years ago
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