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frez [133]
3 years ago
10

How should this combustion reaction be balanced? C5H8+O2 + CO2 + H20

Chemistry
2 answers:
Genrish500 [490]3 years ago
8 0

Answer:

They would equally balance because of the chemicles.

Explanation:

CaHeK987 [17]3 years ago
3 0
They would equally balance because of the chemical
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Climate is the period of time average weather condition which occurs at a place

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A) Head to tail joining of monomers. :) (confirmed correct answer, I took the test)

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How does the Bohr model show you an atom has become an ion?
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If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai
Rama09 [41]

We are given –

  • Final velocity of car is, v= 0
  • Initial velocity of car is, u= 100 km/hr
  • Time taken, t is = 3 minutes or 180 sec

Here–

\qquad\pink{\bf \longrightarrow  Initial\:  velocity =  100 \:km/hr}

\qquad\sf \longrightarrow  Initial\:  velocity = \dfrac{ 100 \times 1000}{3600} \:m/s

\qquad\pink{ \bf \longrightarrow  Initial\:  velocity = 27.78\: m/s}

Now –

\qquad____________________________

\qquad\purple{\bf \longrightarrow  Acceleration  = \dfrac{Final\: Velocity -Initial  \:Velocity }{Time}}

\qquad\purple{\bf \longrightarrow  Acceleration  = \dfrac{v -u}{t}}

\qquad\sf \longrightarrow  Acceleration = \dfrac{(0- 27.78)}{1800}

\qquad\sf \longrightarrow  Acceleration =\cancel{ \dfrac{- 27.78}{1800}}

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6 0
2 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
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