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Allisa [31]
2 years ago
11

A certain gas is found in the exhaust of automobiles and power plants. Consider a 1.53 L sample of a gas at a pressure of 5.6 x

10' Pa. If the pressure is changed to 1.5 × 10 Pa at a constant temperature, what will be the new volume of the gas?​
Chemistry
1 answer:
Olegator [25]2 years ago
6 0

The new volume of the gas that has an initial pressure of 5.6 x 10⁵ Pa is 5.7L. Details about volume of the gas can be found below.

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

The volume of a gas can be calculated using the Boyle's law equation as follows:

P1V1 = P2V2

Where;

  • P1 = initial pressure
  • P2 = final pressure
  • V1 = initial volume
  • V2 = final volume

1.53 × 5.6 × 10⁵ = 1.5 × 10⁵ × V2

8.568 × 10⁵ = 1.5 × 10⁵V2

V2 = 8.57 × 10⁵ ÷ 1.5 × 10⁵

V2 = 5.7L

Therefore, the new volume of the gas that has an initial pressure of 5.6 x 10⁵ Pa is 5.7L.

Learn more about volume at: brainly.com/question/12357202

#SPJ1

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1. A 9.941 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 14.57 grams of CO2 an
salantis [7]

Answer:

1) empirical formula = CH2O ; the molecular formula = C2H402

2) empirical formule = C3H404 = Molecular formula

Explanation:

CxHyOz  → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 14.57 grams of CO2 there is: (12g Carbon / 44g CO2) * 14.57 g CO2 = 3.9736 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom  This is 2g H/ 16g O

In 5.966 g of water there are: [2 g H / 18 g H2O] * 5.966 g H2O = 0.6629 g H

⇒The original sample had 9.941 g of Sample - 3.9736 g of C - 0.6629 g of H =  5.3045 g of O

2) Calculate number of moles

C: 3.9736 g / 12.0 g/mol = 0.3311 mol

H: 0.6629 g / 1.0 g/mol = 0.6629 mol

O: 5.3045g / 16.0 g/mol =  0.3315 mol

3) Now we should divide each number of mole by the smallest number (0.3311) to find the proportion of the elements

C: 0.3311/ 0.3311= 1

H: 0.6629 / 0.3311 = 2

O: 0.3315/ 0.3311  = 1

This gives us the empirical formula of CH2O

4) Calculate the mass of the empirical formula

Molar mass of Carbon = 12g /mole

Molar mass of Hydrogen = 1g /mole

Molar mass of Oxygen = 16g /mole

mass of the empirical formule = 12 + 2*1 + 16 = 30g

5) Calculate the molecular formula

mass of molecular formule / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

60 / 30 = 2 = n

If we multiply CH2O by 2 we'll get: C2H4O2

If we control this by calculating the molar mass:

2*12 + 4*1.01 + 2*16 = 60.04 g/mole

Then, the molecular formula is C2H4O2

CxHyOz  → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 17.97 grams of CO2 there is: (12g Carbon / 44g CO2) * 17.97 g CO2 = 4.90 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom  This is 2g H/ 16g O

In 4.905 g of water there are: [2 g H / 18 g H2O] * 4.905 g H2O = 0.545 g H

⇒The original sample had 14.16 g of Sample - 4.90 g of C - 0.545 g of H =  8.715g of O

2) Calculate number of moles

C: 4.90 g / 12.0 g/mol = 0.4083 mol

H: 0.545 g / 1.0 g/mol = 0.545 mol

O: 8.715g / 16.0 g/mol =  0.5447 mol

3) Now we should divide each number of mole by the smallest number (0.4083) to find the proportion of the elements

C: 0.4083/ 0.4083= 1

H: 0.545 / 0.4083 =1.33

O: 0.5447/ 0.4083  = 1.33

We should multiply everything by 3

This gives us the empirical formula of C3H4O4

4) Calculate the mass of the empirical formula

Molar mass of Carbon = 12g /mole

Molar mass of Hydrogen = 1g /mole

Molar mass of Oxygen = 16g /mole

mass of the empirical formule = 3*12 + 4*1 + 4*16 = 104

5) Calculate the molecular formula

mass of molecular formule / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

104 / 104 = 1 = n

This means the empirical formula = molecular formula = C3H4O4

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