l T-Mobile LTE 10:17 PM Questions (show your work for the quanitative problems)
ID: 104685 • Letter: L
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l T-Mobile LTE 10:17 PM Questions (show your work for the quanitative problems) I. Would this experiment still work if sagar was substituted for salt in Part A7 Explain your Exp:Pnical and Chmical Change 2 la Part A, the mass of beaker A·51.72 the mass of flask B.4081 g;the mass of beaker C ither paper-52.25A mixture of naphthalene, salt and sand is added to beaker A. Its mass is now $5.52 g. The beaker and its contents are heated uncil all the white stuff sublimes. The beaker and ies contents now weigh $4.92& Distilled water (20 mL) is added to the beaker and the contents are soirred for several minutes. The contents are Filtered into a 50-mL Erleameyer flask (B) and rinsed with distilled water with the rinse water added to flask B. The contents of the filter and the filler paper are placed in beaker C. Rask B and Beaker C are placed in an oven and dried for a week, first for two days at 90 "Cand then for 5 days at 120°C, The mass of flask B and its contents is now 42.35, and the mass of beaker C and its conneers is now 5382 g· Calculate the % recovered for each component (naphthalene, sal, sand) and the total %recovered as a % of the original mass 3. How would your results for each component affected (ow, high, or no effect) for each of the %salt, and % sand) in Part A be Explain your answer is each case The stirrer was removed without rinsing it off with distilled water and taking the rinse water and retarning it to beaker A a) b) The stadent misread the first weighing of beaker A (lise ) the true mass was actually 51.82 s not 51.72Explanation / Answer
2.
Mass of beaker A = 51.72 g
Mass of beaker A + nap+ salt+sand= 55.52 g
Mass of nap+ salt+sand= 55.52-51.72
= 3.8 g
White part that sublime = napthalene
mass of napthalene = 55.52 -54.92 = 0.6 g
mass of beaker C after drying = 53.82
Mass of sand = 53.82 - (mass of beaker C + filter paper)
= 53.82 - 52.25
= 1.57 g
Mass of beaker B after drying = 42.35
Mass of salt = 42.35 - 40.81
=1.54 g
Total mass recovered = 1.54 + 1.57 + 0.6
= 3.71
Inital mass = 3.8 g
% recovered = 3.71/3.8 * 100
= 97.6 % recovery
4.
2 H2O(l) -> 2 H2 (g) + O2 (g)
When an electric current is passed through liquid water (H2O), it changes the water into two gases—hydrogen and oxygen. The molecules of water break apart into individual atoms. Since this involve changes in properties of indiviual elements it is a chemical reaction
36 g H2O means 2 moles of H2O
2 moles of H2O give 2 moles og H2 and 1 mole of O2
1 mole O2 weighs 32 g
2 moles of H2 weigh 4 g
This is also in accordance to conservation of mass pricniple (36 = 32 + 4 )
1 molecule H2O contain 2 molecules of H and one molecule of O
total indiviual molecule in 1 molecule of H2O = 3
%O = (1/3) * 100
= 33.33 %
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