Which of the following laboratory procedures best illustrates the law of conservation of mass?
2 answers:
<u>Answer: </u>The correct statement is reacting 12 g of C with 32 g of to form 44 g of
and unused reactants.
<u>Explanation:</u>
Law of conservation of mass states that mass can neither be created nor be destroyed, but it can only be transformed from one form to another. The total mass in the reaction remains constant.
From the given options:
- 1. Reacting 12 g of C with 32 g of
to form 44 g of
and unused reactants
The equation for this is given by:
Mass on the reactant side = (12 + 32)g = 44 g
Mass on the product side = 44 g
Hence, this equation follows Law of Conservation of mass.
- 2. Weighing 100 g of Cu powder and 100 g of Fe filings
As, there is no equation formed, so there will be no chemical reaction and will not follow, law of conservation of mass.
- 3. Heating 100 g of
to produce 56 g of CaO
The equation for this is given by:
Mass on the reactant side = 100 g
Mass on the product side = (56 + 0)g = 56g
Hence, this equation does not follow Law of Conservation of mass.
- 4. Dissolving 10 g of NaCl in 100 g of water
As, salt is getting dissolved in water and hence, it is a physical change and will not follow law of conservation of mass.
Therefore, the correct statement is reacting 12 g of C with 32 g of to form 44 g of
and unused reactants.
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Why?
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For this question, we want to go from liters of solution to mass of HCl, and the conversion factor is laid out as follows:
Have a nice day!
#LearnwithBrainly
<h2>Answers to the whole assignment:</h2>
**Check for proof photos at the bottom.**
**Answers are in bold.**
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A student places three ice cubes in a beaker and allows them to partially melt. If she measures the temperature of the water in the beaker, what will she most likely observe? Explain your answer.
She will see that the temperature of the water is 0 degrees Celsius and it stays at that temperature until all the ice cubes completely melt.
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Use the table on the right to calculate each required quantity. Record values to 3 significant digits.
First answer: 35.8 kJ
Second answer: 871 g
Third answer: Fe
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An iron sphere with a mass of 75.00 g is heated to a temperature of 385.0°C. It is then placed in a beaker containing 150.0 g H2O at 100.0°C.
First answer: 339 kJ
Second answer: 9.02 kJ
Third answer: No
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A sample of water has a mass of 100.0 g. Calculate the amount of heat required to change the sample from ice at –45.00°C to water vapor at 175.0°C. Relevant constant values are given to the right.
First answer: 325 kJ
Second answer: boiling the water
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Identify the correct description for each part labeled on the diagram.
A: ✔ heating solid
B: ✔ melting
C: ✔ heating liquid
D: ✔ boiling
E: ✔ heating gas
F: ✔ melting point
G: ✔ boiling point
Where are the particles closest together?
✔ A
<h2>Explanation:</h2>When ice partially melts, the surrounding water remains at 0° Celsius. This is because ice melting only involves the water molecules overcoming intermolecular forces, or forces holding molecules together. The water molecules themselves aren't increasing in kinetic energy when turning into liquid. Since temperature measures kinetic energy, the temperature in this case, remains at 0°.
To find how much energy it takes to cause a phase change in a substance, you must find the number of moles, and determine whether to use ΔHvap and ΔHfus. ΔHvap is used to show how much energy is involved in evaporation and condensation, and ΔHfus is the same but used for melting and freezing. To find number of moles of a given substance, use the formula: mass of given substance divided by atomic mass. Both numbers have to use the same mass unit. Then, multiply number of moles by ΔHvap or ΔHfus to get kJ.
In a heating curve graph, the flat lines represent no change in energy, or the phase changes. The steep lines are change in energy, which are either heating or cooling. The bottom line is a heating/cooling solid. A heating/cooling solid is the coldest phase, which is why the line is at the bottom.
Here are photos of Edge to make things easier.