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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 aqueous solution of sucrose (C12H22O11C12H22O11) is prepared by dissolving 7.6330 gg in sufficient deionized water to form a
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Answer:

The answer to your question is 0.64 M

Explanation:

Data

Sucrose C₁₂H₂₂O₁₁ mass = 7.633 g

volume = 25 ml

Molarity = ?

Process

1.- Calculate the molar weight of Sucrose

C₁₂H₂₂O₁₁ = (12 x 12) +(22 x 1) + (11 x 16)

                = 144 + 22 + 176

                = 342 g

2.- Calculate the moles of sucrose

                   342 g ------------------ 1 mol

                    7.633 g ---------------  x

                     x = (7.633 x 1) / 342

                    x = 0.0223 moles

3.- Calculate the molarity

Molarity = moles / volume (L)

Molarity = 0.0223 / 0.035

Molarity = 0.64

6 0
3 years ago
In a lab, a student dissolves 6.9 g of sodium chloride (NaCl) in 125 g of water (H2O).
kati45 [8]
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Write the balanced standard combustion reaction for the c5h7on3s2
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Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

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Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

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2 years ago
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