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BigorU [14]
3 years ago
11

A sample of gas has a volume of 20.0 liters at 22.0° C. If the pressure remains constant, what is the volume at 100.0° C?

Chemistry
2 answers:
miv72 [106K]3 years ago
6 0
The problem applies Charles' law since constant pressure with varying volume and temperature are given. Assuming ideal gas law, the equation to be used is \frac{ V_{1} }{ T_{1} }=\frac{ V_{2} }{ T_{2} }. We make sure the temperatures are expressed in Kelvin, hence the given added with 273. The volume 2 is equal to 25.2881 liters.
padilas [110]3 years ago
4 0

Answer: The volume at 100.0^0C is 25.3 L

Explanation:

To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=20.0L\\T_1=22.0^oC=(22.0+273)K=295.0K\\V_2=?\\T_2=100.0^oC=(100.0+273)K=373.0K

Putting values in above equation, we get:

\frac{20.0}{295.0K}=\frac{V_2}{373.0}\\\\V_2=25.3L

The volume at 100.0^0C is 25.3 L

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Consider a possible triatomic molecule containing a nitrogen, oxygen, and fluorine atoms for the following questions.
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Answer:

a) nitrogen

b) nitrogen =5

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Fluorine =7

Explanation:

Usually, if we have two or more elements in a compound, the central atom in the compound is the atom having the least value of electro negativity.

If we consider fluorine, oxygen and nitrogen; nitrogen is the least electronegative of the trio hence it should be the central atom of the triatomic molecule.

The number of valence electrons on the valence shell of each atom is shown below;

nitrogen =5

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Fluorine =7

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Answer:

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Explanation:

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2Al(s)+Fe2O3(s)−→−heatAl2O3(s)+2Fe(l) 2Al(s)+Fe2O3(s)→heatAl2O3(s)+2Fe(l) If 26.1 kg Al26.1 kg Al reacts with an excess of Fe2O3
Dmitry_Shevchenko [17]

Answer : The mass of Al_2O_3 produced will be, 49.32 Kg

Explanation : Given,

Mass of Al = 26.1 Kg  = 26100 g

Molar mass of Al = 26.98 g/mole

Molar mass of Al_2O_3 = 32 g/mole

First we have to calculate the moles of Al.

\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{26100g}{26.98g/mole}=967.38moles

Now we have to calculate the moles of Al_2O_3.

The balanced chemical reaction is,

2Al+Fe_2O_3\rightarrow Heat+Al_2O_3+2Fe

From the balanced reaction we conclude that

As, 2 moles of Al react to give 1 mole of Al_2O_3

So, 967.38 moles of Al react to give \frac{967.38}{2}=483.69 moles of Al_2O_3

Now we have to calculate the mass of Al_2O_3.

\text{Mass of }Al_2O_3=\text{Moles of }Al_2O_3\times \text{Molar mass of }Al_2O_3

\text{Mass of }Al_2O_3=(483.69mole)\times (101.96g/mole)=49317.0324g=49.32Kg

Therefore, the mass of Al_2O_3 produced will be, 49.32 Kg

3 0
3 years ago
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