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shtirl [24]
3 years ago
12

During the recrystallization of a solid, the sample is dissolved in hot solvent then filtered while hot. The filtrate is allowed

to cool and the mixture is filtered a second time. What effect will result if the filtrate solution is cooled too rapidly? Small impure crystals will form. Small pure crystals will form. The maximum amount of the compound as a purified solid will be obtained. Large impure crystals will form.
Chemistry
1 answer:
Inessa [10]3 years ago
5 0

Answer:

<em>If the filtrate solution is cooled too rapidly, Small pure crystals will form.</em>

Explanation:

The crystals form by two processes, nucleation and particle growth. Thus the size of the crystals is determined by the predominance between one or the other. In order to obtain the lower number of nucleus  and the larger crystals it is necessary to cool the solution slowly. If the solution  is cooled too rapidly there is no time to the particles growth and the result will be small crystals. But also, a fast cooling avoid the impurities in the crystals.

   

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Copper roofs on houses form patina (copper carbonates) over the course of years due to the reaction with oxygen, carbon dioxide
ale4655 [162]

Answer:

The answer is "Choice A".

Explanation:

In this question, the choice A is wrong because when the copper panels on houses produce a patina (copper oxide) through occur as a consequence of reactions from oxygen and water in the air, these reaction does not have low activation energy because lower activation energy that is stored in a faster response rate.

5 0
3 years ago
Is turning something anhydrous endo or exothermic?
seraphim [82]
It is endothermic because heat energy is supplied to make the water of crystallization evaporate and make the compound anhydrous.
3 0
3 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
Copy the lists of measurements shown below. Pay close attention to the units that follow each number. List 1: 150 mL 11 mL 200 m
stich3 [128]

a) cross out 11mL from list 1, 2 mL from list 2, and 998 cm3 from list 3.

b) circle 200 mL from list 1, 801 mL from list 2, and 1 L from list 3.

7 0
3 years ago
Read 2 more answers
A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. the boiling point of this solution was dete
Katarina [22]
We will use boiling point formula:

ΔT = i Kb m 

when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35

and Kb is the boiling point constant =5.03

and m = molality 

i = vant's Hoff factor

so by substitution, we can get the molality:

1.35 = 1 * 5.03 * m

∴ m = 0.27

when molality = moles / mass  Kg

           0.27 = moles /  0.015Kg

∴ moles = 0.00405 moles

∴ The molar mass = mass / moles
                               = 2 g /  0.00405 moles 
                               = 493.8 g /mol
5 0
3 years ago
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