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diamong [38]
4 years ago
5

PLEASE ANSWER ARE THESE CORRECTTTTTT

Chemistry
2 answers:
Harman [31]4 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]4 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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8 0
2 years ago
A certain liquid X has a normal freezing point of 7.60 °C and a freezing point depression constant K= 6.90 °C-kg-mol. Calculate
Dmitry [639]

<u>Answer:</u> The freezing point of solution is -5.11°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of sodium chloride follows:

NaCl(aq.)\rightarrow Na^{+}(aq.)+Cl^-(aq.)

The total number of ions present in the solution are 2.

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (NaCl) = 7.57 g

M_{solute} = Molar mass of solute (NaCl) = 58.44 g/mol

W_{solvent} = Mass of solvent (liquid X) = 350.0 g

Putting values in above equation, we get:

\text{Molality of }NaCl=\frac{7.57\times 1000}{58.44\times 350.0}\\\\\text{Molality of }NaCl=0.370m

To calculate the depression in freezing point, we use the equation:

\Delta T=iK_fm

where,

i = Vant hoff factor = 2

K_f = molal freezing point depression constant = 6.90°C/m

m = molality of solution = 0.370 m

Putting values in above equation, we get:

\Delta T=2\times 6.90^oC/m.g\times 0.370m\\\\\Delta T=5.11^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

\Delta T = 5.11 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

5.106^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-5.11^oC

Hence, the freezing point of solution is -5.11°C

7 0
4 years ago
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