how many grams of Urea on heating yield 10 ^22 molecules of biuret by the reaction 2CO(NH2)2=H2N-CO-NH-CO+NH3

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Shailendra Kumar Sharma · Answer

now to produce 10^22 molecules we have to use 2X10^22 molecules of Urea i.e. 2X10^22 /(6.02X10^23) moles So weight will be (2/60.2)*60.06= 1.99g

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how many grams of Urea on heating yield 10 ^22 molecules of biuret by the reaction 2CO(NH2)2=H2N-CO-NH-CO+NH3

how many grams of Urea on heating yield 10 ^22 molecules of biuret by the reaction 2CO(NH2)2=H2N-CO-NH-CO+NH3

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how many grams of Urea on heating yield 10 ^22 molecules of biuret by the reaction 2CO(NH2)2=H2N-CO-NH-CO+NH3

how many grams of Urea on heating yield 10 ^22 molecules of biuret by the reaction 2CO(NH2)2=H2N-CO-NH-CO+NH3

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