4. Combustion of 1.110 g of a gaseous hydrocarbon yields 3.613 g COnand 1.109 g

Question

4. Combustion of 1.110 g of a gaseous hydrocarbon yields 3.613 g COnand 1.109 g H:0, and no other products 0.288 g sample of the hydrocarbon occuples a volume of 131 ml at 24.8 C and 753 mmg. Determine both the empirical and molecular formula. Answer 5. A cubic piece of uranium metal (specific heat capacity . 0.1171/cd at 200.0 C is dropped into 1.00 L of deuterium oxide ("heavy water, specific heat capacity 4.211 yca) at 25.5 C. The final temperature of the the densities of uranium (19.05 /cm) and deuterium oxide (1.11 g/cm*), what is the edge length of the cube of uranlum? Answer 6. A 0.167 g of an unknown acid requires 27.8 mlL of o.100 M NaOH to titr analysis of the acid gives the following percentages by mass: 40.00% GS, 71% H and 53.29% molecular formula, molar mass and Lewis structure of the unknown acid. ate to the equlvalence point. Elemental Determine the Answer ion is composed of C, N and an unknown element X. The skeletal Lewis structure of the 7. A polyatomic ion is X-C-Nr. The ion Xa-has an electron configuration of 4s 3d "4p. What is ele ent x? Knowing the of x, complete the Lewis structure of the polyatomic ion, including all important resonance structures Answers: in a hydrogen atom relaxes to the ns 4 level, emitting ight at 114 THE. What is the value of n for the Aswer

Explanation / Answer

4. no of mol of CO2 produced = 3.613/44 = 0.0821 mol

no of mol of C = 0.0821 mol

no of mol of H2O produced = 1.109/18 = 0.0616 mol

no of mol of H produced = 0.0616*2 = 0.1232 mol

mass of O present = 1.110 -((0.0821*12)+0.1232*1) = 0.0016 g

no of mol of O = 0.0016/16 = 0.0001 mol

simplest ratio,

C = 0.0821/0.0001 = 821

H = 0.1232/0.0001 = 1232

O = 0.0001/0.0001 = 1

molarmass of hydrocatbon(M) = wRT/PV

     = (0.288*0.0821*(24.8+273.15))/((753/760)*0.131)

     = 54.3 g/mol

the question is not clear

5. heat lost by uranium = heat gained by D2O

    m*s*DT = m*s*DT

   m*0.117*(200-28.5) = (1*10^3*1.11)*4.211*(28.5-25.5)

   m = mass of uranium metal = 698.84 g

   volume of cube = mass/density = 698.84/19.05 = 36.7 cm^3

edgelength(a) = V^(1/3)

                = (36.7)^(1/3)

                = 3.323 cm

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