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prisoha [69]
3 years ago
5

An inflated balloon contains X air molecules.After some time the volume of the balloon is found to be the half at the same tempe

rature and pressure when a few air molecules are expelled out. a)How many molecules will be there in the balloon now? b) Which is the gas law associated with this?
Chemistry
1 answer:
astra-53 [7]3 years ago
8 0

<u>Answer:</u>

(a): The balloon will contain \frac{X}{2} number of molecules now.

(b): The gas law associated with this is Avogadro's law.

<u>Explanation:</u>

Avogadro's law states that the volume of the gas is directly proportional to the number of moles of the gas at constant pressure and temperature

Mathematically,

V\propto n                   (At constant pressure and temperature)

OR

\frac{V_1}{n_1}=\frac{V_2}{n_2}               .....(1)

where,

V_1\text{ and }n_1 = Initial volume and number of moles

V_2\text{ and }n_2 = Final volume and number of moles

We are given:

V_2=\frac{V_1}{2}

Putting values in equation 1, we get:

\frac{V_1}{n_1}=\frac{V_1}{2\times n_2}\\\\n_2=\frac{n_1}{2}

(a):

If n_1 number of moles of gas contains X number of molecules

So, \frac{n_1}{2} number of moles of gas will contain = \frac{X}{n_1}\times {n_1}{2}=\frac{X}{2} number of molecules

Hence, the balloon will contain \frac{X}{2} number of molecules now.

(b): The law used is Avogadro's law.

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Air is made of mostly oxygen, nitrogen, argon, carbon dioxide, and small amounts of other substances. Air is a ___________.
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Chemical change.

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KatRina [158]

Explanation:

As the given chemical reaction equation is as follows.

      2Cl_{2}(g) + 7O_{2}(g) + 130 kcal \rightarrow 2Cl_{2}O_{7}(g)

Also, it is given that for 2 moles the energy required is 130 kcal. This means that energy required for 1 mole is calculated as follows.

                   1 mole = \frac{130 kcal}{2}

                               = 65 kcal

Hence, energy required for 7 moles will be calculated as follows.

              Energy required = 7 \times 65 kcal

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Thus, we can conclude that energy required to produce 7.00 mol Cl_{2}O_{7} on the basis of given reaction is 455 kcal.

3 0
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A 30.0-g piece of metal at 203oC is dropped into 400.0 g of water at 25.0 oC. The water temperature rises to 29.0 oC. Calculate
oee [108]

The specific heat of the metal is 2.4733 J/g°C.

Given the following data:

  • Initial temperature of water = 25.0°C
  • Final temperature of water = 29.0°C
  • Mass of water = 400.0 g
  • Mass of metal = 30.0 g
  • Temperature of metal = 203.0°C

We know that the specific heat capacity of water is 4.184 J/g°C.

To find the specific heat of the metal (J/g°C):

Heat lost by metal = Heat gained by water.

Q_{metal} = Q_{water}

Mathematically, heat capacity or quantity of heat is given by the formula;

Q = mc\theta

<u>Where:</u>

  • Q is the heat capacity or quantity of heat.
  • m is the mass of an object.
  • c represents the specific heat capacity.
  • ∅ represents the change in temperature.

Substituting the values into the formula, we have:

30(203)c = 400(4.184)(29 \; -\; 20)\\\\6090c = 1673.6(9)\\\\6090c = 15062.4\\\\c = \frac{15062.4}{6090}

Specific heat capacity of metal, c = 2.4733 J/g°C

Therefore, the specific heat of the metal is 2.4733 J/g°C.

Read more: brainly.com/question/18691577

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