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Mademuasel [1]
3 years ago
7

what is the best course of action if solid material remains in the flask after the heating step of recrystallization

Chemistry
1 answer:
lys-0071 [83]3 years ago
7 0

Answer:

filter the hot mixture.

Explanation:

Solid is stayed undissolved since the arrangement is gotten super saturated. On the off chance that solid molecule is available recrysallization won't happen in this way we need expel the solid molecule by filtarion in hot condition itself . Subsequently, arrangement become totally homogenous and recrysallization item will shaped by moderate cooling

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Write the empirical formula of at least four binary iconic compounds that could be formed from the following ions:
Nat2105 [25]

Explanation:

<h2><em>Explain</em></h2><h2><em>Explainexplain</em></h2><h2><em>Explainexplainexplain</em></h2><h2><em>Explainexplainexplainexplain</em></h2><h2><em>Explainexplainexplainexplainexplain</em></h2>
7 0
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

6 0
3 years ago
Is gamma rays blocked very easily
Leto [7]
No, since they are the strongest type of ray only elements that are dense can block them.
5 0
3 years ago
Help with this please
Len [333]

Answer:

12

Explanation:

There are 4 sulfur atoms in SO4

4×3=12

This means that it turns into 3×(SO4)

=3SO4

6 0
3 years ago
A chemist has to prepare 250.0 mL of a 0.300 M Na2SO4(aq) solution. What mass, in grams, of sodium sulfate (formula mass 142.05
Sedbober [7]

The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g

<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
  • Volume = 250 mL = 250 / 1000 = 0.25 L
  • Molarity = 0.3 M
  • Mole of Na₂SO₄ =?

Mole = Molarity x Volume

Mole of Na₂SO₄ = 0.3 × 0.25

Mole of Na₂SO₄ = 0.075 mole

<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
  • Molar mass of Na₂SO₄ = 142.05 g/mol
  • Mole of Na₂SO₄ = 0.075 mole
  • Mass of Na₂SO₄ =?

Mass = mole × molar mass

Mass of Na₂SO₄ = 0.075 × 142.05

Mass of Na₂SO₄ = 10.65 g

Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.

Learn more about molarity:

brainly.com/question/15370276

6 0
2 years ago
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