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irakobra [83]
3 years ago
10

When placed in a 200-mL square box, a substance has a spherical shape and has a volume of 75 mL. When placed in a 200-mL circula

r tube, the same amount of that same substance has a spherical shape and a 75-mL volume. In which of the following states is the substance?a. solid
b. liquid
c. gas​
Chemistry
1 answer:
Anastasy [175]3 years ago
3 0

Answer:

a. Solid

Explanation:

<em>Solid substances do not change size or shape according to their containers</em>. In other words, solid substances keep their size and shape.

<em>Liquid and gas substances take the shape of their container</em>, so any of them would be square-shaped when put in a square box, and they would be spherical shaped when put in a spherical container.

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I don’t get this pls help
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Explanation:

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The amount of heat required to raise the temperature of a sample of water is referred to as the
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3 years ago
Place the butane lighter in the sink or tub and let it rest there until needed. why do we soak the lighter in the water bath?
Genrish500 [490]

Answer:

To regulate the gas pressure in the lighter tank and avoid build-up of pressure.  

Explanation:

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6 0
3 years ago
A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som
jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of -6.4^oC and a molal freezing point depression constant K_f=3.96^oC.kg/mol. A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at -13.6^oC . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of glycine that can be dissolved is 1.3\times 10^2g

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m

OR

\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}           ......(1)

where,

Freezing point of pure solvent = -6.4^oC

Freezing point of solution = -13.6^oC

i = Vant Hoff factor = 1 (for non-electrolytes)

K_f = freezing point depression constant = 3.96^oC/m

m_{solute} = Given mass of solute (glycine) = ?

M_{solute} = Molar mass of solute (glycine) = 75.07 g/mol

w_{solvent} = Mass of solvent = 950. g

Putting values in equation 1, we get:

-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g

Hence, the mass of glycine that can be dissolved is 1.3\times 10^2g

5 0
3 years ago
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