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denpristay [2]
3 years ago
9

In the technique of recrystallization, what is the 'mother liquor'? The mother liquor is the product in its liquid form. The mot

her liquor is always the non-limiting reactant in the reaction. The mother liquor is the filtrate. The mother liquor always refers to ethanol.
Chemistry
1 answer:
Yakvenalex [24]3 years ago
3 0

In the technique of recrystallization "the mother liquor is the filtrate".

<u>Option: </u>C

<u>Explanation:</u>

The portion of a solution remaining over after crystallization is understood as a mother liquor. It is found in chemical reactions that include sugar refining. It is the liquid produced by filtration of the crystals. The residual liquid, once the crystals have been extracted out as the mother liquor will include a portion of the initial solution as estimated at that temperature by its solubility as well as any unfiltered contaminants. Second and third crystal crops can then be collected from the mother's liquor.

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When metals, non metals,and metallics are composed according to luster, which is best description of the appearance of metalloid
Nitella [24]

Answer:

Metalloids are metallic-looking brittle solids that are either semiconductors or exist in semiconducting forms, and have amphoteric or weakly acidic oxides. Typical nonmetals have a dull, coloured or colourless appearance; are brittle when solid; are poor conductors of heat and electricity; and have acidic oxides.

Explanation:

5 0
2 years ago
If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce
Natali5045456 [20]
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)

65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)

=0.0282669621 g of O2 left over
5 0
3 years ago
Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

6 0
2 years ago
Please help!
Len [333]
It will slowly rotten and turn brown
7 0
3 years ago
3.2 g of a gas at STP occupies a volume of 2.24 L. The gas is
Ronch [10]

Answer:

Oxygen

Explanation:

4 0
2 years ago
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