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Crank
3 years ago
12

Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of CO2 and 11.5 g of H2O. What is the empirical formula for the h

ydrocarbon?
Show work please?
Chemistry
1 answer:
cluponka [151]3 years ago
3 0
<span>C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5</span>
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8 0
3 years ago
Describe in your own words, in terms of particle movement and energy, what occurs when a liquid is heated to its boiling point.
Evgen [1.6K]

Particulate movement and energy increases when a liquid is heated to its boiling point. Explanation: When a liquid is heated to its boiling point, the form of the liquid changes. ... The particles with cinematic energy begin to move more randomly. So we can say that the particle movement and the energy grow.

3 0
3 years ago
9.86 x 10²⁸ O-atoms require what volume (L) N₂O₂ at STP?
jolli1 [7]
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8 0
3 years ago
The polymer formed from the monomer ch2=ch–cn is select one:
podryga [215]
Answer:
             None of the given options show polymer made up of H₂C=CH-CN (Acrylonitrile).

Explanation:
                   Acrylonitrile (H₂C=CH-CN) which is a monomer on self linkage results in a large chain polymer called as Polyacrylonitrile.
                   The structure of Polyacrylonitrile is as follow,

                                             --(H₂C-CHCN-)n--

Where n shows the number of Acrylonitrile units joined together in the formation of Polyacrylonitrile. This polymerization reaction can take place by different mechanisms including free radical mechanism, acid catalyzed addition or base catalyzed addition reaction.

The polymerization is shown below,

5 0
3 years ago
The rate of a certain reaction is given by the following rate law: rate Use this information to answer the question below. At a
AleksAgata [21]

Answer:

Initial rate of the reaction when concentration of hydrogen gas is doubled will be 3.2\times 10^6 M/s.

Explanation:

N_2+3H_2\rightarrow 2NH_3

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Initial rate of the reaction = R = 4.0\times 10^5 M/s

R = k\times [N_2][H_2]^3

4.0\times 10^5 M/s=k\times [N_2][H_2]^3

The initial rate of the reaction when concentration of hydrogen gas is doubled : R'

[H_2]'=2[H_2]

R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3

R'=8\times k\times [N_2][H_2]^3

R'=8\times R=8\times 4\times 10^5 M/s=3.2\times 10^6 M/s

Initial rate of the reaction when concentration of hydrogen gas is doubled will be 3.2\times 10^6 M/s.

7 0
3 years ago
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