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aleksandr82 [10.1K]
3 years ago
8

Jacob has a bucket of ice. He covers the bucket with a lid and sets it in the sun. The ice melts into liquid water. How does the

mass of the ice compare with the mass of the water?
Chemistry
2 answers:
Alekssandra [29.7K]3 years ago
6 0

Answer:

The mass of the ice remains the same in comparison with the mass of water.

Explanation:

As the bucket of ice melted with the help of the sun, the mass of ice remains the same as the mass of ice because there was not any evaporation or adding of water to the bucket, the only thing that happened was a change in the state of matter from solid to a liquid called fusion.

vekshin13 years ago
4 0
Assuming that none of the liquid evaporates, the mass of the ice would be the same as the mass of the water because no chemical change occurred, only a phase change occurred.

Hope this helps
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An experiment was designed to test the hypothesis that peanuts have more energy than a chip. The experiment determines calorimet
Svet_ta [14]

Answer:

b Different amounts of food samples were used.

Explanation:

The mass of the two samples needs to be the same in order for the test to be accurate.

4 0
1 year ago
How many molecules are in 100 g of C6H120,?*​
Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

6 0
3 years ago
The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.376 g sample of e
hoa [83]

Answer:

The answer to your question is: C₄H₁₀O

Explanation:

Data

          CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

                      CxHyOz  + O₂ ⇒   CO₂  +  H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

                      44 g of CO₂ --------------  12 g of C

                      3.268 g of CO₂  --------    x

                         x = (3.268 x 12) / 44

                        x = 0.891 g of Carbon

                       12 g of carbon -----------  1 mol

                       0.891 g of C     ----------   x

                       x = (0.891 x 1) / 12

                       x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

                      18 g of water --------------- 2 g of H

                      1.672 g of H₂O ------------  x

                      x = (1.672 x 2) / 18

                      x = 0.186 g of hydrogen

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

                      0.186 g of H       ------------  x

                      x = (0.186 x 1) / 1

                      x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

                           = 0.299 g of O₂

Moles of Oxygen

                             16 g of Oxygen ---------------- 1 mol

                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

                             x = 0.019 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon         0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

4 0
3 years ago
6. What is the molecular formula for this compound? The compound's empirical formula and
igomit [66]

Explanation:

A compound's empirical formula tells you the smallest whole number ratio ,The molar mass tells you what the total mass of one mole

6 0
2 years ago
What is the purpose of constructing a calibration curve
stellarik [79]

Answer:

Calibration curves are used to understand the instrumental response to an analyte, and to predict the concentration of analyte in a sample.

5 0
3 years ago
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