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Radda [10]
3 years ago
14

if a gas sample has a pressure of 761 mmhg at 0.00°c by how much does the temperature have to decrease to lower the pressure to

759 mmhg?​
Chemistry
1 answer:
Nady [450]3 years ago
6 0
<h3>Answer:</h3>

272.43 K or -0.718°C

<h3>Explanation:</h3>

We are given;

The initial pressure,P1 as 761 mmHg

Initial temperature, T1 as 0.00°C which is equivalent to 273.15 K

Final pressure as 759 mmHg

We are required to calculate the final temperature;

According to pressure law, the pressure of a gas and absolute temperature are directly proportional at constant volume.

That is; Pα T

Therefore, at varying pressure and temperature,

\frac{P1}{T1}=\frac{P2}{T2}

To get final temperature;

T2=\frac{P2T1}{P1}

T2=\frac{(759 mmHg)(273.15K)}{761 mmHg}

T2=272.43K

Therefore, the final temperature will be 272.43 K or -0.718°C

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How many moles of electrons are transferred when 2.0 moles of aluminum metal react with excess copper(II) nitrate in aqueous sol
mojhsa [17]

Moles of electrons:

The moles of electrons that are transferred are 12F

A balanced equation:

2 moles of Aluminium metal react with excess copper(II) nitrate.

2Al + 3Cu{(NO_3)}_2  \rightarrow 2Al{(NO_3)}_3 +3 Cu

Given:

Moles of Aluminium = 2

As Aluminium goes from 0 to +3 oxidation state

Al \rightarrow Al^{3+} + 3e^-

And copper goes from +2 to 0

Cu^{2+} + 2e^-\rightarrow Cu

On balancing the number of electrons we get:

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So, 12F moles of electrons are transferred.

Learn more about Faraday's Law here,

brainly.com/question/27985929

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4 0
2 years ago
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3 years ago
24H2S + 16HNO3 3Sg+ 16NO + 32H2O
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3 0
3 years ago
If 4.12 l of a 0.850 m-h3po4 solution is be diluted to a volume of 10.00 l, what is the concentration of the resulting solution?
statuscvo [17]

The process in which the concentration of the solution is lessened  by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.

Formula is given by:

C_{1}V_{1}=C_{2}V_{2}   (1)

where,

C_{1} is the initial concentration

V_{1} is the initial volume

C_{2} is the final concentration

V_{2} is the final volume

Now,

C_{1} = 0.850 M

V_{1} = 4.12 L

C_{2}  =?

V_{2} = 10.00 L

Substitute the give values in formula (1),

0.850 M\times 4.12L=C_{2}\times 10.00 L

C_{2} =\frac{0.850 M\times 4.12L}{10.00 L}

= 0.3502 M

Thus, the final concentration of theH_{3}PO_{4} solution = 0.3502 M












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3 years ago
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