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ANTONII [103]
3 years ago
13

After a recrystallization, a pure substance will ideally appear as a network of __________. If this is not the case, it may be w

orthwhile to reheat the flask and allow the contents to cool more _________
Chemistry
1 answer:
Stells [14]3 years ago
4 0

Answer:

Large crystals

Slowly

Explanation:

Recrystallization is a procedure used to purify an impure compound in a solvent. The concept used here is that the solubility of most of the solids increases with increase in temperature.

Recrystallization is also called as method of fractional crystallization.

After a recrystallization, a pure substance will ideally appear as a network of <u>large crystals</u>. If this is not the case, it may be worthwhile to reheat the flask and allow the contents to cool more <u>slowly</u>.

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Answer:

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Explanation:

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2 years ago
A water bath in a physical chemistry lab is 1.85m long, 0.810m wide and 0.740m deep. If it is filled to within 2.57 in from the
jenyasd209 [6]

Answer:

The volume of water in water bath is 1,011 Liters.

Explanation:

Length of the water bath, L = 1.85 m

Width of the water bath, W= 0.810  m

Height of the water bath ,H= 0.740 m

Height of the water in water bath, h= 0.740 m - 2.57 inches

1 m = 39.37 inch

= 0.740 m - \frac{2.57}{39.37} m = 0.6747 m

Volume of the water in bath = L × W × h

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1 m^3=1000 L

1.011 m^3=1.011\times 1000 L=1,011 L

The volume of water in water bath is 1,011 Liters.

6 0
4 years ago
If 250.0 g of water at 30.0 °C cool to 5.0 °C, how many kilojoules of energy did the water lose?
Y_Kistochka [10]

Answer:

-26.125 kj

Explanation:

Given data:

Mass of water = 250.0 g

Initial temperature = 30.0°C

Final temperature = 5.0°C

Amount of energy lost = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 5.0°C - 30.0°C

ΔT = -25°C

Specific heat of water is 4.18 j/g.°C

Now we will put the values in formula.

Q = m.c. ΔT

Q = 250.0 g × 4.18 j/g.°C × -25°C

Q = -26125 j

J to kJ

-26125 j ×1 kj /1000 j

-26.125 kj

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