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Yuri [45]
3 years ago
6

How would you prepare 10 ml of a 0.25% m/v hcl solution if 1% m/v hcl was available? how much 1% m/v hcl is needed? how much dis

tilled water is used?
Chemistry
1 answer:
Fed [463]3 years ago
8 0

To solve this we use the equation, 

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

1% x V1 = 0.25% x 10 mL

V1 =2.5 mL

Therefore, you will need to have 2.5 mL of the 1% HCl solution and 7.5 mL of distilled water. In mixing the two liquids, you should remember that the order of mixing would be acid to water. So, you use a 10 mL volumetric flask . Put small amount of distilled water and add the 2.5 mL of HCl solution. Lastly, dilute with distilled water up to the 10 mL mark.

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3 years ago
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Answer:

–2733.4 KJ

Explanation:

The balanced equation for the reaction is given below:

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From the balanced equation above,

1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.

Finally, we shall determine the enthalpy change (ΔH) produced by the reaction of 2 moles of C₂H₅OH. This can be obtained as follow:

From the balanced equation above,

1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.

Therefore, 2 moles of C₂H₅OH will react to produce enthalpy change (ΔH) of = 2 × −1366.7 = –2733.4 KJ.

Thus, enthalpy change (ΔH) obtained is –2733.4 KJ

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Explanation: Picture attached.

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