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lara31 [8.8K]
3 years ago
14

Which of the following compounds has the most deshielded protons?A) CH3ClB) CH3IC) CH3BrD) CH4

Chemistry
1 answer:
Korolek [52]3 years ago
6 0

Given :

Some compounds :

A)\ CH_3Cl\ B)\ CH_3I\ C)\ CH_3Br\ D)\ CH_4 .

To Find :

Which of the following compounds has the most deshielded protons .

Solution :

Deshielded means nucleus whose chemical shift has been increased due to removal of electron density, magnetic induction, or other effects .

In simple words deshielding means the ability to shift protons .

Now , among Cl , I , Br and H . Cl is the most electron negative .

Therefore , deshielding will be more in CH_3Cl .

Hence , this is the required solution .

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If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
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Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

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3 years ago
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