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प्रश्न
24 cc Marsh gas (CH4) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.
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उत्तर
This experiment supports Gay Lussac’s law of combining volumes.
According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.
\[\ce{CH4 + 2O2 → CO2 + 2H2O}\]
1V 2V 1V 0V
24 cc 48 cc 24 cc
1 vol. of CH4 requires O2 = 2 vols.
∴ 24 cc of CH4 requires O2 = 2 × 24 = 48 cc
Similarly, 24 cc of CH4 produce CO2 = 24 cc
∴ Vol. of O2 unused = 106 − 48 = 58cc
∴ Vol. of mixture on cooling = unused oxygen + CO2 formed
= 58 + 24
= 82 cc
संबंधित प्रश्न
Propane burns in air according to the following equation:
C3H8 + 5O2 → 3CO2 + 4H2O
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?
When gases react their volumes bear a simple ratio to each other, under the same conditions of temperature and pressure. Who proposed this gas law?
Propane burns in air according to the following equation:
\[\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\]
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?
What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions? (assuming oxygen is `1/5`th of air)
\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
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\[\ce{2NO + O2 → 2NO2}\]
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\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
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- If the volume of A is actually 5.6 dm3 at STP, calculate the number of molecules in the actual Volume of D at STP (Avogadro's number is 6 × 1023).
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