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Novay_Z [31]
2 years ago
9

Complete combustion of 1.00 g of the hydrocarbon pagodane gives 3.38 g carbon dioxide. what is the empirical formula of pagodane

?
Chemistry
1 answer:
Blababa [14]2 years ago
6 0
Combustion of any hrdrocarbon yields carbon dioxide and water such that the hydrogen and carbon are derived from the hydrocarbon. 
1 mole of carbon dioxide has a mass out of which 12 g is the mass of carbon, therefore, 3.38 g of carbon will contain  (3.38 × 12)/44 = 0.922 g of carbon.
Thus, since the hydrocarbon has a mass of 1 g, then the mass of hydrogen will be (1-0.922 g) = 0.078 g.
To get the empirical formula we divide the mass of each element by the atomic mass to get the number of moles.
Carbon= 0.922/12 = 0.0768 moles
Hydrogen = 0.078/1 = 0.078 moles
Then we get the ratio of the moles of carbon : hydrogen
                                                      =   0.0768 : 0.078
                                                         =         1 : 1.016
                                                          ≈        1:   1
Therefore the empirical formula of the hydrocarbon will be CH


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The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required t
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2 years ago
What is the total number of sp2-hybridized carbon atoms present in the fluorophore used in the experiments
Mandarinka [93]

Answer:

9

Explanation:

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2 years ago
A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimete
N76 [4]

Answer:

The mass of water = 219.1 grams

Explanation:

Step 1: Data given

Mass of aluminium = 32.5 grams

specific heat capacity aluminium = 0.921 J/g°C

Temperature = 82.4 °C

Temperature of water = 22.3 °C

The final temperature = 24.2 °C

Step 2: Calculate the mass of water

Heat lost = heat gained

Qlost = -Qgained

Qaluminium = -Qwater

Q = m*c*ΔT

m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)

⇒with m(aluminium) = the mass of aluminium = 32.5 grams

⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C

⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C =  -58.2 °C

⇒with m(water) = the mass of water = TO BE DETERMINED

⇒with c(water) = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C

32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9

-1742.1 = -7.95m

m = 219.1 grams

The mass of water = 219.1 grams

8 0
2 years ago
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4vir4ik [10]

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hence the molar mass of the gas comes out to be 106.6 g

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