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igomit [66]
3 years ago
9

In a recrystallization (a technique that you will encounter later in the semester), a solid is dissolved in a solvent and later

the solvent is removed. if a mp of the sample is taken while the sample is still moist with solvent, what effect would that have on the mp of the sample?
Chemistry
1 answer:
grandymaker [24]3 years ago
7 0

In a recrystallization (a technique that you will encounter later in the semester), a solid is dissolved in a solvent and later the solvent is removed. if a mp of the sample is taken while the sample is still moist with solvent then melting point will not come out to be accurate, as solvent is still present in the compound and that results in the variation of the melting point as the residual solvent will serves as impurities , and due to the presence of these impurities, melting point vary and does not come to be accurate .

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Answer:

Lithium diisopropylamide [(CH3)2CH]2NLi.

Explanation:

Lithium diisopropylamide (commonly abbreviated LDA) is a chemical compound with the molecular formula [(CH3)2CH]2NLi. It is used as a strong base and has been widely accepted due to its good solubility in non-polar organic solvents and non-nucleophilic nature. It is a colorless solid, but is usually generated and observed only in solution. It was first prepared by Hamell and Levine in 1950 along with several other hindered lithium diorganylamides to effect the deprotonation of esters at the α position without attack of the carbonyl group.

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ehidna [41]
A is 1. Each letter corresponds to what place they are in the alphabet
4 0
3 years ago
How many milliliters of 0.1 m hcl is required to react with 0.15g sodium carbonate (na2co3?
docker41 [41]
1) Write the balanced chemical equation

     2HCl + Na2 CO3 ----------> 2NaCl + H2CO3

2) Write the molar ratios:

    2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3

3) Convert 0.15g of sodium carbonate to number of moles

3a) Calculate the molar mass of Na2CO3

Na: 2 * 23 g/mol = 46 g/mol

C: 12 g/mol =

O: 3 * 16 g/mol = 48 g/mol

molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol

3b.- Calculate the number of moles of Na2CO3

# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3

4) Calculate the number of moles of HCl from the molar proportion:

[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl

5) Calculate the volume of HCl from the definition of Molarity

Molarity, M = # moles / volume in liters

=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters

0.028 liters * 1000 ml / liter = 28 ml.

Answer: 28 mililiters of 0.1 M HCl.
7 0
3 years ago
High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen. (a) Determine t
Lemur [1.5K]

Answer:

Concentration of Flourine = 24.756%

Explanation:

Given that :

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.

the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

At standard conditions , the atomic weight of the these compounds are as follows:

Carbon = 12.01 g/mol

Chlorine = 35.45 g/mol

Fluorine = 19.00 g/mol

Hydrogen = 1.008 g/mol

Oxygen = 16.00 g/mol

The chemical formula for polyethylene = (CH₂ - CH₂)ₙ

Therefore, for two carbons, there will be 4 hydrogens;

i.e

(CH₂ - CH₂)₂

( C₂H₄ - C₂H₄ )

Suppose the number of original hydrogen = 4moles

number of moles of Flourine F = 12% of 4

= 0.12 × 4

= 0.48 mol

∴ the number of remaining moles of Hydrogen is:

= 4 - 0.48

= 3.52 moles

number of moles of Carbon = 2 moles

∴ the mass of flourine F = number of moles of F × molar mass of F

= 0.48 × 19

= 9.12

The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)

= 9.12 + 3.52 + 24

= 36.64

Concentration of Flourine = (mass of flourine/total mass) × 100

Concentration of Flourine = (9.12/36.84 ) × 100

Concentration of Flourine = 0.24756 × 100

Concentration of Flourine = 24.756%

3 0
3 years ago
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