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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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Answer:

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Explanation:

w = Weight of an object = 100 N

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m = Mass of an object

Weight is given by

w=mg\\\Rightarrow m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{100}{9.81}\\\Rightarrow m=10.19\ \text{kg}

Mass of the object is 0.19 kg

g = 2\ \text{m/s}^2

m = 4 kg

w=mg\\\Rightarrow w=4\times 2\\\Rightarrow w=8\ \text{N}

Weight of the object is 8 N.

m = 200 g

w=mg\\\Rightarrow w=0.2\times 9.81\\\Rightarrow w=1.964\ \text{N}

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m = 6kg

w=mg\\\Rightarrow w=6\times 9.81\\\Rightarrow w=58.86\ \text{N}

The force of the object is 58.86 N.

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