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VladimirAG [237]
2 years ago
10

A drop of water with a mass of 0.48 g is vaporized at 100 ∘C and condenses on the surface of a 55- g block of aluminum that is i

nitially 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 ).)
Express the temperature in Celsius to two significant figures.
Chemistry
1 answer:
bagirrra123 [75]2 years ago
7 0

The final temperature in Celsius of the metal block is 49°C.

<h3>How to find the number of moles ?</h3>

Moles water = \frac{\text{Given mass}}{\text{Molar Mass}}

                     = \frac{0.48\ g}{18\ \text{g/mol}}

                     = 0.0266 moles  

                   

Heat lost by water = 0.0266 mol x 44.0 kJ/mol

                                = 1.17 kJ

                                = 1170 J           [1 kJ = 1000 J]

Heat lost = Heat gained

Heat gained by aluminum = 1170 J  

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

1170 + 1242 = 49.7 T  

T = 48.5°C (49°C at two significant figures)

Thus from the above conclusion we can say that The final temperature in Celsius of the metal block is 49°C.

Learn more about the Moles here: brainly.com/question/15356425

#SPJ1

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Cytokinesis happens differently for plant and animal cells. Both separate cytoplasm between two new daughter cells. However, whi
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Explanation:

Cytokinesis -

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The process of cytokinesis is different for animal cell and plant cell.

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What is the empirical formula of a compound composed of 3.25% hydrogen ( H ), 19.36% carbon ( C ), and 77.39% oxygen ( O ) by ma
Montano1993 [528]

Answer:

The answer to your question is C₂HO₃

Explanation:

Data

Hydrogen = 3.25%

Carbon = 19.36%

Oxygen = 77.39%

Process

1.- Write the percent as grams

Hydrogen = 3.25 g

Carbon = 19.36 g

Oxygen = 77.39 g

2.- Convert the grams to moles

                     1 g of H ----------------- 1 mol

                   3,25 g of H -------------  x

                     x = (3.25 x 1) / 1

                     x = 3.25 moles

                    12 g of C ---------------- 1 mol

                     19.36 g of C ----------  x

                     x = (19.36 x 1) / 12

                     x = 1.61 moles

                     16g of O --------------- 1 mol

                     77.39 g of O ---------  x

                      x = (77.39 x 1)/16

                      x = 4.83

3.- Divide by the lowest number of moles

Carbon = 3.25/1.61 = 2

Hydrogen = 1.61/1.61 = 1

Oxygen = 4.83/1.61 = 3

4.- Write the empirical formula

                        C₂HO₃

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