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3 years ago

PLEASE ANSWER ARE THESE CORRECTTTTTT

Chemistry

2 answers:

Harman [31]3 years ago

6 0

Left Frame

The answer you have chosen is not correct. You have selected a physical property. Color is something that will not change as long as nothing chemical is done to the sample. If it is just going to sit there at room temperature in a flask and be admired, (particularly if the flask has a cork in it), then whatever you see is a physical property. The only one that isn't is E. When you burn something, there is a change. One chemical turns into something else. If you burn gasoline (for your car) then C8H18 + 13.5 02 ==> 8CO2 + 9H20 is the result. That gasoline is not easily reversible. It has changed into something else (Water and Carbon Dioxide). That's Chemical.

Middle Frame

I've never encountered these terms before. Your two examples of extensive are correct. I'm not sure about the chemical reaction. I think it is intensive, but I wouldn't bet the farm on it. A chemical reaction is going to take place no matter how much is used. You could have too much sample or too much acid, but what happens does not depend on the amount. I think you are correct.

Answer: you have placed these in the right place.

Right frame.

All four are in the right category. Very nicely done.

Naddik [55]3 years ago

3 0

1. The answer is E.

2. I agree with your answers.

3. I also agree with your answers

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If there's just some barium put in an aqueous solution, then it should be something like this.
It's a mixture of a solution and an insoluble solid, so the easiest way to go is through filtration. (Also, I'm assuming the barium is cut into very small chips.)

In a line, simply filter the solution using a folded filter paper in a funnel, collect the residue in a beaker or flask, rinse it with distilled water and let it dry. (Or simply filtering it could be enough, depending on how far your teacher wants you to go.)

Stuff needed:
>filter paper (for separating the solid from the solution)
>funnel (to hold the filter paper)
>beaker or flask (to hold the filtrate)
>distilled water (to rinse the solid)
>spatula (to scoop up the solid)

Procedure:
>Fold filter paper and line the funnel with it. Place the funnel in the flask or beaker.
>Pour solution in. Then add water (I think using tap water might be fine in this case, but you can use distilled water if you'd like) to wash out the container with the solution of any solid you may have not gotten in the first try. Alternatively, you could use a spatula to spoon it onto the filter paper.
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>Take out the filter paper, open it up and let it dry.

This can be used in real life in many occasions. For example, when you make tea, you need to filter the leaves out. Or when you cook the pasta, you put it in a sieve to separate the pasta from the water. Or when you fish using fishing nets, you "filter" the fish from the water.

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Read 2 more answers

Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 9.045×10−2 −mg sample of menthol i

Ket [755]

Answer:

Empirical Formula = C₁₀H₂₀O

Solution:

Data Given:

Mass of Menthol = 9.045 × 10⁻² mg = 9.045 × 10⁻⁵ g

Mass of CO₂ = 0.2546 mg = 0.0002546 g

Mass of H₂O = 0.1043 mg = 0.0001043 g

Step 1: Calculate %age of Elements as;

%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

%C = (0.0002546 ÷ 9.045 × 10⁻⁵) × (12 ÷ 44) × 100

%C = (2.814) × (12 ÷ 44) × 100

%C = 2.814 × 0.2727 × 100

%C = 76.73 %

%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

%H = (0.0001043 ÷ 9.045 × 10⁻⁵) × (2.02 ÷ 18.02) × 100

%H = (1.153) × (2.02 ÷ 18.02) × 100

%H = 1.153 × 0.1120 × 100

%H = 12.91 %

%O = 100% - (%C + %H)

%O = 100% - (76.73% + 12.91%)

%O = 100% - 89.64%

%O = 10.36 %

Step 2: Calculate Moles of each Element;

Moles of C = %C ÷ At.Mass of C

Moles of C = 76.73 ÷ 12.01

Moles of C = 6.3888 mol

Moles of H = %H ÷ At.Mass of H

Moles of H = 12.91 ÷ 1.01

Moles of H = 12.7821 mol

Moles of O = %O ÷ At.Mass of O

Moles of O = 10.36 ÷ 16.0

Moles of O = 0.6475 mol

Step 3: Find out mole ratio and simplify it;

C H O

6.3888 12.7821 0.6475

6.3888/0.6475 12.7821/0.6475 0.6475/0.6475

9.86 19.74 1

≈ 10 ≈ 20 1

Result:

Empirical Formula = C₁₀H₂₀O₁

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3 years ago