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Sonbull [250]
2 years ago
8

Convert 1176 mmHg to kPa

Chemistry
1 answer:
OLEGan [10]2 years ago
8 0

Answer:

 156.8kPa

Explanation:

The problem here is to convert mmHg to kPa;

We have been given:

            1176mmHg and the problem is to convert to kPa;

  1000Pa = 1kPa

               1 mmHg  =  133.322Pa

               1176mmHg will give  1176 x 133.322  = 156787.1Pa

To kPa;

             156.8kPa

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Which of the following two combinations of reactants is more appropriate for the preparation of p-nitrophenyl phenyl ether? Fluo
VladimirAG [237]

Answer:

p-fluoronitrobenzene and sodium phenoxide is more appropriate

Explanation:

An ipso substitution is required to form p-nitrophenyl phenyl ether.

For this ipso substitution, an alkoxide anion needs to attack as a nucleophile at the carbon atom attached to fluorine atom and thereby substitute that F atom.

p-nitrophenoxide is an weak nucleophile as compared to phenoxide due to presence of electron withdrawing resonating effect of nitro group at para position.

p-fluoronitrobenzene is a good choice for nucleophilic attack by alkoxide anion as compared to fluorobenzene due to higher positive charge density at carbon atom directly attached to F atom. Higher positive charge density arises due to presence of electron withdrawing resonating effect og nitro group at para position.

So, p-fluoronitrobenzene and sodium phenoxide is more appropriate

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3 years ago
An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

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