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ehidna [41]
3 years ago
8

Consider a solution that is 0.05 M HCl. Your goal is to neutralize 1 L of this solution (i.e. bring the pH to 7). You also have

a solution that is 5 M NaOH. What volume of this solution should you add to the HCl solution, to neutralize it? Provide your answer in units of liters (L).
Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0

Answer:

The volume of NaOH required is - 0.01 L

Explanation:

At equivalence point ,

Moles of HCl = Moles of NaOH

Considering :-

Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}

Given  that:

Molarity_{NaOH}=5\ M

Volume_{NaOH}=?\ L

Volume_{HCl}=1\ L

Molarity_{HCl}=0.05\ M

So,  

Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}

0.05\times 1=5\times Volume_{NaOH}

Volume_{NaOH}=\frac{0.05\times 1}{5}=0.01\ L

<u>The volume of NaOH required is - 0.01 L</u>

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The combustion of 0.590 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
Alenkinab [10]

Answer:

2.943 °C temperature change from the combustion of the glucose has been taken place.

Explanation:

Heat released on combustion of Benzoic acid; :

Enthaply of combustion of benzoic acid = 3,228 kJ/mol  

Mass of benzoic acid = 0.590 g

Moles of benzoic acid = \frac{0.590 g}{122.12 g/mol}=0.004831 mol

Energy released by 0.004831 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004831 mol=15.5955 kJ=15,595.5 J

Heat capacity of the calorimeter = C  

Change in temperature of the calorimeter = ΔT = 2.125°C

Q=C\times \Delta T

15,595.5 J=C\times 2.125^oC

C=7,339.05 J/^oC

Heat released on combustion of Glucose: :

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=1.400 g

Moles of glucose =\frac{1.400 g}{180.16 g/mol}=0.007771 mol

Energy released by the 0.007771 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.007771 mol=21.6030 kJ=21,603.01 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

21,603.01 J=7,339.05 J/^oC\times \Delta T'

\Delta T'=2.943^oC

2.943 °C temperature change from the combustion of the glucose has been taken place.

8 0
3 years ago
Please help will give brainliest to right answer!!
Nezavi [6.7K]

Answer:

C

Explanation:

NP

3 0
3 years ago
Read 2 more answers
A gas has a pressure of 8.5atm and occupies 24L at 25∘C. What volume (in liters) will the gas occupy if the pressure is increase
Vesna [10]

The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L

From the question given above, the following data were obtained:

Initial pressure (P₁) = 8.5 atm

Initial volume (V₁) = 24 L

Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K

Final pressure (P₂) = 13.5 atm

Final temperature (T₂) = 15 °C = 15 + 273 = 288 K

<h3>Final volume (V₂) =? </h3>

  • The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\\\frac{8.5 * 24}{298}  = \frac{13.5 * V_{2}}{288}\\\\ \frac{204}{298} = \frac{13.5 * V_{2}}{288}\\\\

Cross multiply

298 × 13.5 × V₂ = 204 × 288

4023 × V₂ = 58752

Divide both side by 4023

V_{2} = \frac{58752}{4023}\\\\

<h3>V₂ = 15 L </h3>

Therefore, the final volume of the gas is 15 L

Learn more: brainly.com/question/25547148

3 0
2 years ago
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