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worty [1.4K]
3 years ago
7

Find the initial temperature of Aluminum if 35 g of it produces 781.2 Joules at the final temperature of 44.3 ºC.

Chemistry
1 answer:
Annette [7]3 years ago
6 0

Answer:

T initial =  19.55°C

Explanation:

Given data:

Mass of Al = 35 g

Energy produces = 781.2 J

Final temperature = 44.3°C

Initial temperature = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of Al is 0.902 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

Q = m.c. ΔT

781.2 J = 35 g×0.902 J/g.°C× (T final - T initial)

781.2 J = 35 g×0.902 J/g.°C× (44.3°C -T initial )

781.2 J = 31.57 j/°C× (44.3°C -T initial )

781.2 J/ 31.57 j/°C = (44.3°C -T initial )

24.75°C = (44.3°C -T initial )

24.75° - 44.3°C = -T initial

T initial =  19.55°C

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A solution is made by dissolving 23.5 grams of glucose (C6H12O6) in 0.245 kilograms of water. If the molal freezing point consta
aalyn [17]

Answer:

- 0.99 °C ≅ - 1.0 °C.

Explanation:

  • We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.

<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>

4 0
3 years ago
A solution that contains a large amount of salt and a small amount of water is said to be a _______ solution. A. diffused B. uns
Pachacha [2.7K]

Answer:

D. Concentrated

Explanation:

5 0
3 years ago
Read 2 more answers
Taking advantage of their large differences in pKa values, describe how a mixture of phenol and benzoic acid in diethyl ether so
Marat540 [252]

Answer:

By adding bicarbonate.

Explanation:

The pka of the phenol (C₆H₅OH) is 10 and the pka of the benzoic acid (C₆H₅COOH) is 4, which means that the benzoic acid is a stronger acid than phenol, so if we want to separate phenol from benzoic acid in diethyl ether we need to first use a weak base that will react with benzoic acid and not with the phenol:  

C₆H₅-COOH + HCO₃⁻  ⇄  C₆H₅-COO⁻  +  H₂CO₃

C₆H₅-OH + HCO₃⁻  ⇄  no reaction

The reaction of the benzoic acid with bicarbonate will produce the benzoate ion that will be soluble in the aqueous layer, while the phenol will remain dissolved in the organic layer, so we can separate the two of them by the separation of the two immiscible layers.      

Having the two layers separated, the benzoic acid can be recovered from the aqueous layer by adding HCl:

C₆H₅-COO⁻ + HCl  ⇄  C₆H₅-COOH + Cl⁻

<u>This acid will precipitate from the aqueous solution, and the solid can be isolated by filtration</u>.  

The phenol in the organic layer can be dissolved into an aqueous layer by the adding of a strong base like NaOH:

C₆H₅-OH + OH⁻  ⇄  C₆H₅-O⁻ + H₂O

The phenoxide ion soluble in the aqueous layer can be recovered later by the adding of HCl, which will form the original phenol:

C₆H₅-O⁻ + HCl  ⇄  C₆H₅-OH + Cl⁻  

<u>The precipitated phenol can be isolated by filtration. </u>

This way we can separate a mixture of phenol and benzoic acid in diethyl ether solution.  

I hope it helps you!

6 0
3 years ago
Which two characteristics describe all animals
Maru [420]

All animals can be dangerous and they would fight for their family. (This might be wrong)

5 0
3 years ago
If 2.00 moles of H₂ and 1.55 moles of O₂ react how many moles of H₂O can be produced in the reaction below?
jekas [21]

Answer:

2 mol H₂O

Explanation:

With the reaction,

  • 2H₂(g) + O₂(g) → 2 H₂O(g)

1.55 moles of O₂ would react completely with ( 2*1.55 ) 3.1 moles of H₂. There are not as many moles of H₂, thus H₂ is the limiting reactant.

Now we <u>calculate the moles of H₂O produced</u>, <em>starting from the moles of limiting reactant</em>:

  • 2.00 mol H₂  * \frac{2molH_2O}{2mol H_2} = 2 mol H₂O
5 0
3 years ago
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