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Tamiku [17]
3 years ago
10

You are given three beakers of unknown liquids. One beaker contains pure water. One contains salt water. One contains sugar wate

r. Without tasting the liquids, how could you identify the liquid in each beaker?
Chemistry
2 answers:
Verdich [7]3 years ago
6 0

Answer: The density and electrical conductivity of the solutions can be measured

Explanation:

We could get three beakers and measure the mass of the solution by subtracting the mass of the beaker from mass of solution + beaker. After that, we use a measuring cylinder to determine the volumes of the three solutions. The density is then obtained from:

Density= mass/volume

The density of pure water is always 1gcm-3.

Measurement of electrical conductivity will distinguish between the salt and sugar solutions. The salt solution will have high electrical conductivity. The solution with the highest electrical conductivity is the salt solution while the solution with the least electrical conductivity is the sugar solution. Sugar does not conduct electricity in solution because it does not contain ions.

sergeinik [125]3 years ago
5 0
Trick question: you just stated everything in the beakers.
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4 0
3 years ago
Read 2 more answers
Mg + 2HCl → MgCl2 + H2
deff fn [24]

Answer:

Explanation:

This is a limiting reactant problem.

Mg(s)

+

2HCl(aq)

→

MgCl

2

(

aq

)

+ H

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(

g

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Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).

4.86

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×

1

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g Mg

=

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Determine Moles of 2M Hydrochloric Acid

Convert

100 cm

3

to

100 mL

and then to

0.1 L

.

1 dm

3

=

1 L

Convert

2.00 mol/dm

3

to

2.00 mol/L

Multiply

0.1

L

times

2.00 mol/L

.

100

cm

3

×

1

mL

1

cm

3

×

1

L

1000

mL

=

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2.00 mol/dm

3

=

2.00 mol/L

0.1

L

×

2.00

mol

1

L

=

0.200 mol HCl

Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,

2.01588 g/mol

0.200

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×

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mol H

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mol Mg

×

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g H

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1

mol H

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=

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0.200

mol HCl

×

1

mol H

2

2

mol HCl

×

2.01588

g H

2

1

mol H

2

=

0.202 g H

2

The limiting reactant is

HCl

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0.202 g H

2

under the stated conditions.

pls mark as brainliest ans

7 0
2 years ago
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4 0
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