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Mrac [35]
4 years ago
15

Uppose the reaction between naphthalene and 2-bromo-2-methylpropane did not run to completion, so that the limiting reagents wer

e not consumed. how would the two TLC plates have looked in your first and second run in this case?
A)there wouuld have been two spots in both cases

B)There would have been three spots in both cases

C)There would have been four pots in both cases

D)There would have been no distinct spots but only streaks.
Chemistry
1 answer:
ch4aika [34]4 years ago
3 0

Answer:

C)There would have been four pots in both cases

Explanation:

In an incomplete reaction of naphthalene and 2-bromo-2-methylpropane, we started with 2 spots (one each for the two starting materials) and we should end up with 3 spots If spots will appear clearly, one for naphthalene, one for the alkyl halide and one for the product. However, this is possible to have same number of spots initially and finally when we consider 2 spots each for naphthalene and 2-bromo-2-methylpropane then after reaction we have 2 spots of products as each 1 spot of reactants decreases. Finally making initial 4 and final 2+1+1 =4.

So, the answer is (C)

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A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.
kaheart [24]

The equilibrium constant of the reaction is 282. Option D

<h3>What is equilibrium constant?</h3>

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

Concentration of iodine = 1.600 mol/1.000-L =  1.600M

In this case, we must set up the ICE table as shown;

              Br2(g) + I2(g) ↔ 2IBr(g)

I          0.6            1.6           0

C      -x                -x             +2x

E    0.6 - x         1.6 - x       1.190

If 2x = 1.190

x = 1.190/2

x = 0.595

The concentrations at equilibrium are;

[Br2] = 0.6 -  0.595 = 0.005

[I2] =   1.6 - 0.595 = 1.005

Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

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Learn more about equilibrium constant:brainly.com/question/15118952

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4 0
2 years ago
How man grams of cl2 are consumed to produce 12.0 g of KCl
Korvikt [17]

Answer:

5.71 g

Explanation:

Step 1: Write the balanced equation

2 K + Cl₂ ⇒ 2 KCl

Step 2: Calculate the moles corresponding to 12.0 g of KCl

The molar mass of KCl is 74.55 g/mol.

12.0 g × 1 mol/74.55 g = 0.161 mol

Step 3: Calculate the moles of Cl₂ needed to produce 0.161 moles of KCl

The molar ratio of Cl₂ to KCl is 1:2. The moles of Cl₂ needed are 1/2 × 0.161 mol = 0.0805 mol

Step 4: Calculate the mass corresponding to 0.0805 moles of Cl₂

The molar mass of Cl₂ is 70.91 g/mol.

0.0805 mol × 70.91 g/mol = 5.71 g

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