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Julli [10]
3 years ago
13

Consider the reaction. mc014-1.jpg How many grams of methane should be burned in an excess of oxygen at STP to obtain 5.6 L of c

arbon dioxide? 2.0 g 4.0 g 16.0 g 32.0 g
Chemistry
2 answers:
vodomira [7]3 years ago
8 0

Answer:

the answer is 4 grams methane.

Explanation:

ur welcome

garri49 [273]3 years ago
5 0
The reaction for the combustion of methane can be expressed as follows.
                           CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
                            moles of CO2 = (5.6 L) / (22.4 L/1 mole)
                             moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
                              moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
                                                    = 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane. 
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5)
Kryger [21]

The empirical formula : C₁₂H₄F₇

The molecular formula : C₂₄H₈F₁₄

<h3>Further explanation</h3>

mol C (MW=12 g/mol)

\tt \dfrac{18.24}{12}=1.52

mol H(MW=1 g/mol) :

\tt \dfrac{0.51}{1}=0.51

mol F(MW=19 g/mol)

\tt \dfrac{16.91}{19}=0.89

mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7

Empirical formula : C₁₂H₄F₇

(Empirical formula)n=molecular formula

( C₁₂H₄F₇)n=562 g/mol

(12.12+4.1+7.19)n=562

(281)n=562⇒ n =2

Molecular formula : C₂₄H₈F₁₄

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3 years ago
An oxygen o2 molecule is adsorbed on a patch of surface (see sketch at right). this patch is known to contain 484 adsorption sit
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We have Boltzmann's equation S = k ln W 
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 W = Number of absorption sites
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 At W = 729, Entropy S2 = 1.381 x 10^-23 ln 729 = 9.103 x 10^-23 J/K
 Change of Entropy = S2 - S1 = 0.566 x 10^-23 J/K
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C has 2H 2O and 2H 2O so is balanced.
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Answer:

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