If 40.6 g of hydrocarbon contain 4.21 × 10^23 molecules then the hydrocarbon can be -?

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If 40.6 g of hydrocarbon contain 4.21 × 10^23 molecules then the hydrocarbon can be -?

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Direction (Q. Nos. 21-26) This section contains 2paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given ptions (

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Two containers X Y are present with container X consisting of some mass of He at some temperature while container Y having double the volume as that of container X kept at same temperature containing same mass of H2 gas as the mass of Helium gas. Based on this data the following conditions answer the question that followsAssume sizes of H2 molecule He atom to be same size of Hatom to be half to that of Heatom only bimolecular collisions to be occuring.Condition I: all except one atom of He are stationary in cont. X all molecules of H2 are moving in container Y.Condition II: both containers contain all moving moleculesQ. Assuming condition II to be applicable, if temperature only of container Y is doubled to that of original (causing dissociation of all H2 gas into H gaseous atoms) then, if no. of total collisions per unit volume per unit time in container X is A then, no. of total collisions made by all molecules per unit volume in container Y would be

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Two containers X Y are present with container X consisting of some mass of He at some temperature while container Y having double the volume as that of container X kept at same temperature containing same mass of H2 gas as the mass of Helium gas. Based on this data the following conditions answer the question that followsAssume sizes of H2 molecule He atom to be same size of Hatom to be half to that of Heatom only bimolecular collisions to be occuring.Condition I: all except one atom of He are stationary in cont. X all molecules of H2 are moving in container Y.Condition II: both containers contain all moving moleculesQ. Assuming condition I to be applicable if no. of total collisions occuring per unit time is A in container X then no. of total collisions made by any one molecule in container Y will be

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Direction (Q. Nos. 27-30) This section contains 4questions. when worked out will result in an integer from 0 to 9 (both inclusiv e) Q. For hydrocarbon with molecular formula C5H8, how many acyclic isomers without consecutive pi-bonds are possible? Correct answer is '3'. Can you explain this answer?

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Three flasks of equal volumes contain CH4, CO2 and Cl2 gases respectively. They will contain equal number of molecules if –

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If 40.6 g of hydrocarbon contain 4.21 × 10^23 molecules then the hydrocarbon can be -? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about If 40.6 g of hydrocarbon contain 4.21 × 10^23 molecules then the hydrocarbon can be -? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 40.6 g of hydrocarbon contain 4.21 × 10^23 molecules then the hydrocarbon can be -?.

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