Dear Student, The computation are expressed below, 2NH3+CO2 →NH2-CO-NH2 + H2Omoles of urea = 999 6×10360=16 66×103CnH2n+2 + O2 →nCO2+H2O16 66×103n×[12n+(2n+2)] = 236×103n≅12Hence the molecular formula of hydrocarbon is,C12H26 Regards

8 q pllll

8 q pllll OBJECT of the following has the maximum number of atoms? a) 24 g Of C ( mol. weight = 12 gmol-l) CHEMISTRY 8. 9. b) 48g@smol.weight = 32 gmol-') C) 23 g Of Na ( mol. weight 23 gmoi•') Reaction Of Br2 with Na2C03 in aqueous solution gives sodium bromide and sodium bromate with evolution of C02 gas. The number of sodium bromide molecules involve in the balance Chemical equation is 3. If we consider that 1/6 in place 1/12, mass of carbon atom is taken to be the relative atomic mass unit. the mass of one mole of substance will ow many moles Of magnesium phosphate, contain 0.25 mole of oxygen atoms? The density (in gmL•l) of a 3.60 M suiÅ6uric acid solution that is 29 % H2S04 (molar mass: 01-1) by mass will be A 5.2 molal aqueous solution of methyl alcohol CH30H is supplied. What is the mole fraction of nnethyl alcohol in the;olution? • Calculate the number of moles of nitrogen dioxide NOZ that could be prepared from 0.35 mol of nitrogen oxide and 0.25 mol Of Oxygen. 2NO(g) + 02tg) 2N02(9) Urea (H2NCONH2) is manufactured by passing C02(g, through ammonia solution followed by crystallisation. C02 for the above reaction is prepared by combustion of hydrocarbon. If ¯ —combustion of 236 kg of a saturated hydrocarbon (CnH2n.2) produces as much C02 as required for production of 999.6 kg urea then what is the molecular formula of hydrocarbon? The concentration of bivalent lead ions in a sample of polluted water that also contains nitrate ions is determined by adding solid sulphate (MB = 142 gmol•l) to exactly SOO mL of water. Calculate the molarity of lead ions if 0.35" of sodium sulphate was needed for complete recipitation of lead ions as sulphate. Sulphuric acid reacts with sodium hydroxide as follows: H2S04 + 2NadH —Y NazSOÅ + 2H20 When 1 L of 0.1M sulphuric acid solution is allowed to react with IL of 0.1M sodium hydroxide o/ution, what is the amount of sodium sulphate formed and its molarity in the solution? Jf hasmog'obin (molecular weight = 67200) contains 0.33% of iron by weight, then what is the ber of iron atoms present in one molecule of haemoglobin?

Dear Student,

The computation are expressed below,

2 N H_{3} + C O_{2} \to N H_{2} - C O - N H_{2} + H 2 O m o l e s o f u r e a = \frac{999.6 \times 10^{3}}{60} = 16.66 \times 10^{3} C_{n} H_{2 n + 2} + O_{2} \to n C O_{2} + H_{2} O \frac{16.66 \times 10^{3}}{n} \times [12 n + (2 n + 2)] = 236 \times 10^{3} n ≅ 12 H e n c e t h e m o l e c u l a r f o r m u l a o f h y d r o c a r b o n i s, C_{12} H_{26}

Regards.

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