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Nonamiya [84]
2 years ago
8

A gas has a pressure of 853.0 millibars at a temperature of 29.0 °C. If the volume is unchanged but the temperature is increased

to 85.0 °C,
what is the new pressure of the gas?
=
Chemistry
1 answer:
Nitella [24]2 years ago
8 0

The new pressure of the gas that initially have a pressure of 853.0 millibars at a temperature of 29.0 °C is 1011.17 millibars. Details about pressure can be found below.

<h3>How to calculate pressure?</h3>

The pressure of a given gas can be calculated using the following formula:

P1/T1 = P2/T2

Where;

  • P1 = initial pressure = 853.0 millibars
  • P2 = final pressure = ?
  • T1 = initial temperature = 29°C + 273 = 302K
  • T2 = final temperature = 85°C + 273 = 358K

853/302 = P2/358

358 × 853 = 302P2

305374 = 302P2

P2 = 305374 ÷ 302

P2 = 1011.17 millibars

Therefore, the new pressure of the gas that initially have a pressure of 853.0 millibars at a temperature of 29.0 °C is 1011.17 millibars.

Learn more about pressure at: brainly.com/question/15175692

#SPJ1

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Explanation:

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Atoms of noble gases are generally inert because..
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B. their outer electron levels are filled

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6 0
3 years ago
Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI
NeTakaya

Answer :

AgI should precipitate first.

The concentration of Ag^+ when CuI just begins to precipitate is, 6.64\times 10^{-7}M

Percent of Ag^+ remains is, 0.0076 %

Explanation :

K_{sp} for CuI is 1\times 10^{-12}

K_{sp} for AgI is 8.3\times 10^{-17}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

CuI\rightleftharpoons Cu^++I^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][I^-]

1\times 10^{-12}=0.0079\times [I^-]

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Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgI\rightleftharpoons Ag^++I^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][I^-]

8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M

[Ag^+]=6.64\times 10^{-7}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{6.64\times 10^{-7}}{0.0087}\times 100

Percent of Ag^+ remains = 0.0076 %

8 0
4 years ago
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