"Finding the relative atomic mass of lithium."

Determination of the intercourse atomic aggregative of atomic subroutine 3 Introduction: For this investigation I pass on determine the Relative atomic kettle of fish (Ar) by victimization two different modes. In the first method I bequeath dissolve a percentage of lithium of a known mass in water, I result thence collect the atomic number 1 gas produced, which rump be utilise to betoken the relative atomic mass of Lithium. In the mho method I will titrate the resulting rootage of lithium hydroxide with a known concentration of hydrochloric acid, this can as well be used to calculate the relative atomic mass of lithium. Readings: *100 cm3 of water *37.47 g of lithium plus the remain ice rink *37.41 g of Watch glass *122 cm3 of atomic number 1 gas collected Calculations from method 1 Moles of hydrogen produced: Assuming that 1 groine of gas occupies 24000 cm3 at pressure and inhabit temperature! I will now calculate the number of moles of the hydro gen gas produced. I will do that by dividing the mickle of gas produced by 24000cm3 : Moles (n) = Volume (cm) 24000 cm So: n = 122 cm = 0.005083 mol of H2 24000 cm Equation: 2 Li(s) + 2 H2O (L)2LiOH(Ag) + H2 (g) By looking at the Stoichiometry ratio in the reaction I can see that the ratio of Li: H2 is 2:1, therefore the number of moles of Li will be double as the number of moles of H2 in the reaction. So: 0.005083 x 2= 0.

010166 mol of Li tummy of Lithium = Mass of the Lithium plus the watch glass - the mass of the watch glass So: 37.47g - 37.41g = 0.06g ·0.06g of lithium was used Relative at omic mass of lithium: I will now calculate ! the relative atomic mass of the piece of lithium that I used by rearranging a formula. This shove is foreign to me, and I can say that the talking to you use is actually simple and easy to understand. The calculations part is where I lose it. Sorry, likely not much help to you. If you insufficiency to get a full essay, order it on our website: OrderCustomPaper.com

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