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o-na [289]
3 years ago
7

A drop of water weighing 0.48 g is vaporized at 100 ?c and condenses on the surface of a 55-g block of aluminum that is initiall

y at 25 ?c. if the heat released during condensation goes only toward heating the metal, what is the final temperature in celsius of the metal block? (the specific heat capacity of aluminum is 0.903 j/g ?c.
Chemistry
2 answers:
irakobra [83]3 years ago
4 0
<h3><u>Answer;</u></h3>

<em>-49 °C</em>

<h3><u>Explanation and solution;</u></h3>
  • Considering the fact that, the specific heat capacity of aluminum is 0.903 J/g x C, and the heat of vaporization of water at 25 C is 44.0 KJ/mol.  

Moles water = 0.48 g / 18.02 g/mol

                      =0.0266  moles

<em>Heat lost by water</em> = 0.0266 mol x 44.0 kJ/mol

                                 =1.17 kJ => 1170 J  

<em>But heat lost =heat gained</em>

<em>Therefore;</em> Heat gained by aluminium = 1170 J  

        1170 = 55 x 0.903 ( T - 25) = 49.7 T - 1242  

                   1170 + 1242 = 49.7 T  

            T = 48.5 °C ( 49 °C <em>at two significant figures)</em>

<em>Hence</em>, final temperature = 49 °C

Amanda [17]3 years ago
4 0

Hi, the solution would be like this for this specific problem:<span>

(2257 J/g) x (0.48 g) = 1083.36 J </span><span>

(1083.36 J) / (0.903 J/g ∘C) / (55 g) = 21.81°<span>C change </span></span><span>

25°C + 21.81°C = 47°C</span><span>

I hope this helps and if you have any further questions, please don’t hesitate to ask again. </span>

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Answer:

(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.

Explanation:

Without mincing words let's dive straight into the solution to the question.

(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;

The freezing point depression = [ 1 - (-3)]° C = 4°C.

(b). The molality can be Determine by using the formula below;

Molality = the number of moles found in the solute/ solvent's weight(kg).

Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.

(c). The mass of acetone that was in the decanted solution = 11.1 g.

(d). The mass of water that was in the decanted solution = 89.01 g.

(e). 2.4 = x/ 58 × (1000/1000).

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7 0
3 years ago
What would the volume of gas be at 150 c if it had a volume of 693 ml at 45 c​
marissa [1.9K]

Answer:

\boxed{\text{922 mL}}

Explanation:

The pressure is constant, so we can use Charles' Law to calculate the volume.

\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}

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V₁ = 693 mL; T₁ =  45 °C

V₂ = ?;           T₂ = 150 °C

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T₁ = (  45 + 273.15) = 318.15 K

T₂ = (150 + 273.15) = 423.15 K

(b) Calculate the volume

\dfrac{ 693}{318.15} = \dfrac{ V_{2}}{423.15}\\\\2.178 = \dfrac{ V_{2}}{423.15}\\\\V_{2} = 2.178 \times 423.15 = \boxed{\textbf{922 mL}}

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Which is an intensive property of a substance?
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3 years ago
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kakasveta [241]

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Explanation:

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7.(03.01 MC)
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All options are correct except,

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Most of mass comes from nucleus in which protons and neutrons are present.

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e-

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It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties

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Symbol of neutron= n0  

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3 0
3 years ago
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