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MAVERICK [17]
3 years ago
15

List and describe two differences between pure substances and mixtures. Check all that apply. Check all that apply. The composit

ion of a pure substance depends on the source, but the composition of a mixture doesn't depend on the source. The composition of a pure substance is always the same, regardless of the source, but the composition of a mixture can vary. The state of a pure substance is always the same, but the state of a mixture can vary. Mixtures can be separated into their components only by chemical changes; some pure substances can be separated into components by physical change. Mixtures can be separated into their components by physical changes; some pure substances can be separated into components by chemical change.
Chemistry
1 answer:
34kurt3 years ago
3 0

Answer:

The composition of a pure substance is always the same, regardless of the source, but the composition of a mixture can vary.

Mixtures can be separated into their components by physical changes; some pure substances can be separated into components by chemical change.

Explanation:

A chemically pure substance has a well defined and constant composition. The composition of a chemically pure substance remains the same irrespective of its source. Also, the components of a chemically pure substance may be separated by chemical changes.

Mixtures have a variable composition depending on their  respective sources. The composition of a mixture varies with the source of the mixture and mixtures are separated by physical processes.

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A 4 liter solution of bleach with a concentration of 0.5 moles/L is diluted using an extra 4 Liters of water. What is the final
blondinia [14]

Answer:

0.25 mol/L

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 4L

Initial concentration (C1) = 0.5 mol/L

Final volume (V2) = 4 + 4 = 8L

Final concentration (C2) =?

Applying the dilution formula, we can easily find the concentration of the diluted solution as follow:

C1V1 = C2V2

0.5 x 4 = C2 x 8

Divide both side by 8

C2 = (0.5 x 4 )/ 8

C2 = 0.25 mol/L

Therefore the concentration of the diluted solution is 0.25 mol/L

3 0
3 years ago
Why are scientific method important in the study of science
Sergio [31]

Answer: The scientific method. When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.

Explanation:

6 0
3 years ago
What value must be known if the accuracy of an experiment is to be determined​
andrezito [222]

To evaluate the accuracy of a measurement, the measured value must be compared to the correct value. To evaluate the precision of a measurement, you must compare the values of two or more repeated measurements.
7 0
3 years ago
How can one kg of iron melt more ice than 1 kg lead at 100 °C
Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{iron} = Heat obtained from the iron by the ice

ΔQ_{iron} = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{lead} = Heat obtained from the iron by the ice

ΔQ_{lead} = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of  ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

3 0
3 years ago
PLS HALP ASAP points
marishachu [46]

Answer:

50 N ( left)

Explanation:

Given data:

The force on right side = 450 N

The force on left side = 500 N

Net force = ?

Solution:

F (net) = 500 N - 450 N

F (net) =  50 N ( left)

5 0
3 years ago
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