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natita [175]
3 years ago
7

If a sample of air initially occupies 240L at 2 atm how much pressure is required to compress it to 20L at constant temperature

Chemistry
1 answer:
IceJOKER [234]3 years ago
4 0

Answer:

24 atm.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 240 L

Initial pressure (P₁) = 2 atm

Final volume (V₂) = 20 L

Temperature = constant

Final pressure (P₂) =?

The final pressure required, can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

2 × 240 = P₂ × 20

480 = P₂ × 20

Divide both side by 20

P₂ = 480 / 20

P₂ = 24 atm

Thus, the final pressure required is 24 atm.

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4. the heat of reaction

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When NH3(g) reacts with N2O(g) to form N2(g) and H2O(g), 105 kcal of energy are evolved for each mole of NH3(g) that reacts. Wri
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Answer : The balanced chemical equation is,

2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal

Explanation :

Balanced chemical equation : It is defined as the number of atoms of individual elements present on the reactant side must be equal to the number of atoms of individual elements present on product side.

The given unbalanced chemical reaction is,

NH_3(g)+N_2O(g)\rightarrow N_2(g)+H_2O(g)+105kcal

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of individual elements are not balanced.

In order to balanced the chemical reaction, the coefficient 2 is put before the NH_3, the coefficient 3 is put before the N_2O\text{ and }H_2O and the coefficient 4 is put before the N_2.

The energy evolved in this reaction = 105Kcal\times 2=210Kcal

Thus, the balanced chemical reaction will be,

2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal

7 0
3 years ago
What atomic component contributes to many of the trends of elements
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Given 6.98 x 10 4 power grams of iron, calculate the moles of iron present
KiRa [710]

Answer:

1249.88 mol.

Explanation:

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<em>∴ no. of moles of Fe </em>= (6.98 x 10⁴ g)/(55.845 g/mol) = <em>1249.88 mol.</em>

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How much energy (heat) is required to convert 52.0 g of ice at –10.0°C to steam at 100°C? Specific heat of ice 2.09 J/g • °C Spe
dexar [7]

Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ

Explanation:

Using this formular, q = [mCpΔT] and = [nΔHfusion]

The energy that is needed in the different physical changes is thus:

The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:

q = [mCpΔT]

q = 52.0 x 2.09 x 10

q = 1.09 kJ

While from 0°C to 100°C is calculated as:

q = [mCpΔT]

q = 52.0 x 4.18 x 100

q = 21.74 kJ

And for fusion at 0°C is called Heat of fusion and would be given as:

q = n ΔHfusion

q = 52.0 / 18.02 x 6.02

q = 17.38 kJ

And that required for vaporization at 100°C is called Heat of vaporization and it's given as:

q = n ΔHvaporization

q = 52.0 / 18.02 x 40.7

q = 117.45 kJ

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5 0
3 years ago
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