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Lapatulllka [165]
3 years ago
8

Separating Mixtures Checkpoint ( giving brainly )

Chemistry
2 answers:
Setler79 [48]3 years ago
7 0

Answer:

evaporation

Explanation:

Alex Ar [27]3 years ago
4 0
Condensation i think
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Define an ideal gas and explain under which conditions you may reasonably approximate a real gas as an ideal gas. Also mention c
TEA [102]

Answer:

High temperature and low pressure

Explanation:

According to the kinetic molecular theory, gases are composed of small particles called molecules which are in constant motion.

At high temperature and low pressure, gas molecules possess high kinetic energy and move at high velocities hence intermolecular interaction is almost none existent and real gases approach the behavior of ideal gases.

5 0
3 years ago
A certain sample of a liquid has a mass of 42 grams and a volume of 35 centimeters3. What is the density of the liquid?
BabaBlast [244]

Answer:

            1.2 g.cm⁻³

Solution:

Data Given:

                             Mass  =  42 g

                             Volume  =  35 cm³

Formula Used:

                             Density  =  Mass ÷ Volume

Putting values,

                             Density  =  42 g ÷ 35 cm³

                             Density  =  1.2 g.cm⁻³

3 0
3 years ago
Read 2 more answers
How many milliliters of 0.125 M FeCl3 are needed to react with an excess of Na2S to produce 3.75 g of Fe2S3 if the percent yield
Katyanochek1 [597]

Answer:

0.912 mL

Explanation:

3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

FeCl3 is the limiting reactant.

Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles

Hence actual yield of Iron III sulphide = 0.043 moles

Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield

Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide

From the reaction equation,

2moles of iron III chloride produced 1 mole of iron III sulphide

x moles of iron III chloride, will produce 0.057 of iron III sulphide

x= 2× 0.057= 0.114 moles of iron III chloride

But

Volume= number of moles/ concentration

Volume= 0.114/0.125

Volume= 0.912 mL

4 0
3 years ago
Help help help plsss
Olin [163]

Answer:

4.58×10^3

3.72×10^7

Explanation:

3.8×10^2 =380

+ 4.2×10^3=4200

= 4580 = 4.58×10^3

8 0
3 years ago
20. What volume of 0.350M KMnO4 solution must be diluted to prepare 600. mL of
Dafna1 [17]

Answer:

25.7 mL

Explanation:

Step 1: Given data

  • Initial volume (V₁): ?
  • Initial concentration (C₁): 0.350 M
  • Final volume (V₂): 600 mL
  • Final concentration (C₂): 0.150 M

Step 2: Calculate the volume of the initial solution

We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.150 M × 600 mL / 0.350 M

V₁ = 25.7 mL

3 0
3 years ago
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