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oee [108]
4 years ago
15

Look into the picture for the question.​

Chemistry
1 answer:
ira [324]4 years ago
3 0

Answer:

Option D. 1092K

Explanation:

T1 = 273K

V1 = V

P1 = P

V2 = 2V( i.e double of original volume)

P2 = 2P (i.e double of original pressure)

T2 =?

P1V1/T1 = P2V2/T2

PV/273 = 2P x 2V / T2

Cross multiply to express in linear form

PV x T2 = 273 x 2P x 2V

Divide both side by PV

T2 = (273 x 2P x 2V) / PV

T2 = 273 x 2 x 2

T2 = 1092K

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Answer:

Solid

Explanation:

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3 years ago
In a given substance, as the temperature of a substance increases, the volume of that substance _____.
Molodets [167]

Answer:

C: increases

Explanation:

Temperature is directly proportional to volume. If the temperature of a substance increases, the volume also increases. If the temperature decreases the volume also decreases.

According to Charles's law "the volume of fixed mass of gas varies directly as its absolute temperature if the pressure remains constant". Although this is an ideal gas law, it is also applicable to most substances.

As the temperature is raised, the volume also behaves proportionally that way.  

5 0
4 years ago
If Carl buys a 946 ml bottle of rubbing alcohol, how much of the aqueous solution is water?
Darina [25.2K]

The question is incomplete, here is the complete question:

A bottle of rubbing alcohol having aqueous solution of alcohol contains 70% (v/v) alcohol. If Carl buys a 946 ml bottle of rubbing alcohol, how much of the aqueous solution is water?

<u>Answer:</u> The amount of water present in the given bottle of rubbing alcohol is 283.8 mL

<u>Explanation:</u>

We are given:

Volume of bottle of rubbing alcohol = 946 mL

70% (v/v) alcohol solution

This means that 70 mL of rubbing alcohol is present in 100 mL of solution

Amount of water present in solution = [100 - 70] = 30 mL

Applying unitary method:

In 100 mL of solution, the amount of water present is 30 mL

So, in 946 mL of solution, the amount of water present will be = \frac{30}{100}\times 946=283.8mL

Hence, the amount of water present in the given bottle of rubbing alcohol is 283.8 mL

4 0
3 years ago
In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to
Artyom0805 [142]

Answer:

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

Explanation:

The measured pH of the solution = 5.153

C_6H_5OH\rightarrow C_6H_5O^-+H^+

Initially      c

At eq'm   c-x       x  x

The expression of dissociation constant is given as:

K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}

Concentration of phenoxide ions and hydrogen ions are equal to x.

pH=-\log[x]

5.153=-\log[x]

x=7.03\times 10^{-6} M

K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}

K_a=9.34\times 10^{-11}

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

4 0
4 years ago
Part C<br> How did Dr. Tierno find the answer to his question?
matrenka [14]
Sorry I cant I just need some points
7 0
3 years ago
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