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garik1379 [7]
3 years ago
6

Suppose that 1.06 g of rubbing alcohol (C3H8O) evaporates from a 72.0 g aluminum block. If the aluminum block is initially at 25

∘C∘C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C∘C. The heat of vaporization of the alcohol at 25 ∘C∘C is 45.4 kJ/molkJ/mol, the specific heat of aluminum is 0.903 J/g⋅∘CJ/g⋅∘C.
Chemistry
1 answer:
Ksenya-84 [330]3 years ago
4 0

Answer:

The final temeprature of the block is 12.71 °C

Explanation:

Step 1: Data given

Mass of C3H8O = 1.06 grams

Mass of aluminium block = 72.0 °C

The initial temperature of the aluminium block = 25.0 °C

The alcohol vaporizes at 25 ∘C.

The heat of vaporization of the alcohol at 25 ∘C = 45.4 kJ/mol,

The specific heat of aluminum is 0.903 J/g∘C

Step 2: Calculate moles C3H8O

Moles C3H8O = mass C3H8O / molar mass C3H8O

Moles C3H8O = 1.06 grams / 60.1 g/mol

Moles C3H8O = 0.0176 moles

Step 3: Calculate energy

Energy = 45.4 kJ/mol * 0.0176 moles

Energy = 0.799 kJ = 799 J  gained by the alcohol and lost by the Al

Step 4: Calculate the change of temeprature

Q = m*c*ΔT

ΔT = Q / (m*c)

ΔT = 799 J / (72.0*0.903)

ΔT = 12.29 °C

. Step 5: Calculate the final temperature

T2 = 25.0 °C - 12.29 = 12.71 °C

The final temeprature of the block is 12.71 °C

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Answer:The first method to determine the chemical composition of a substance in space was using light. By determining red shift in the observed spectrum of light they could determine the elements they were observing

Explanation:

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Give the complete electronic configuration for oxygen. 1s22s22px22pz2 1s22s22px4 1s22s22py22pz2 1s22s22px22py2 1s22s22px22py12pz
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20. What volume of 0.350M KMnO4 solution must be diluted to prepare 600. mL of
Dafna1 [17]

Answer:

25.7 mL

Explanation:

Step 1: Given data

  • Initial volume (V₁): ?
  • Initial concentration (C₁): 0.350 M
  • Final volume (V₂): 600 mL
  • Final concentration (C₂): 0.150 M

Step 2: Calculate the volume of the initial solution

We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.150 M × 600 mL / 0.350 M

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3 0
3 years ago
A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
sesenic [268]

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

  • 0 mL of HBr is added

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

  • After adding 12 mL of HBr

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

  • Equivalence point

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

  • Titration midpoint (28.2 mL/2)

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

  • When we add 45.1 mL of HBr

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

7 0
3 years ago
What is the relationship between temperature and volume?
Lelu [443]

Answer:

Charles's law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant

The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles's law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle's law)

Explanation:

~Hope this helps

6 0
3 years ago
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