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Serhud [2]
3 years ago
6

Determine the concentration of a solution prepared by diluting 20.0 ml of a 0.200 m rbcl to 250.0 ml. determine the concentratio

n of a solution prepared by diluting 20.0 ml of a 0.200 m rbcl to 250.0 ml. 0.00800 m 0.160 m 0.0160 m 0.0320 m 2.50 m
Chemistry
1 answer:
pogonyaev3 years ago
5 0
When dealing with making diluted solutions from concentrated solutions, we can use the following formula 
c1v1 = c2v2 
where c1 and v1 are the concentration and volume of the concentrated solution respectively.
c2 and v2 are the concentration and volume of the diluted solution respectively 
substituting these values in the above formula,
20 mL x 0.200 M = C x 250.0 mL 
C = 0.0160 M 
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Ethanol and water appear the saame to the naked eye at 25C. Which properties can help a chemist distinguish between? denisty, ma
Oxana [17]

Explanation:

As it is known that there are two types of properties. These are extensive and intensive.

Extensive properties : Properties that depend on the size or amount of system. For example, mass, volume etc.

Intensive properties : Properties that do not depend on the size or amount of system. For example, density, melting point, specific heat capacity etc.

On the basis of these properties water and ethanol are distinguished as follows.

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  • Melting point of water is zero degree celsius whereas melting point of ethanol is -114.1 degree celsius.
  • Specific heat capacity of water is 4.184 J/g ^{o}C whereas specific heat capacity of ethanol is 2.46 J/g ^{o}C.
  • Mass of the given liquids cannot be differentiated because they will keep on changing depending on the quantity required. As mass is an extensive property, therefore, it is difficult to differentiate between the two liquids.

Thus, we can conclude that properties like density, melting point, specific heat capacity can help a chemist distinguish between ethanol and water.

7 0
3 years ago
Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production
Lynna [10]

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

<em>(1) </em>C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

<em>(2) </em>CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

<em>(3) </em>H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

<em>(4) </em>C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

<em>(5) </em>CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = <em>-747.54kJ</em>

<em></em>

Thus <em>ΔH° of the reaction is -747.54kJ</em>

3 0
3 years ago
1. Hydrogen and nitrogen combine to form ammonia. When nitrogen and hydrogen bond, nitrogen pulls the electrons from hydrogen to
UNO [17]

Answer:

did you do it yet?

Explanation:

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