Watch the below videos to answer questions regarding your topic. Your answers must be thorough and include specific detail and diagrams. DO NOT write down information you don’t understand. If need be, research the topic further using your textbook or the world wide web!
Link 1: https://www.youtube.com/watch?v=njGz69B_pUg&feature=youtu.be
Link 2: http://www.youtube.com/watch?v=gmwgJ6bDLHM&safe=active
Link 3: http://www.youtube.com/watch?v=4Gymoej6N2A&safe=active
1. Define all words
2a. P- phase change
2b. C- oxidation
2c. P- Phase change
2d. C- chem. reaction
3. Melting point, density, color, state: physical changes
4. carbon (PS, E); water (PS, C); marble (M, HT, Sus.); diamond (PS, E); ocean water (M, HT, Sus.); #2 pencil (M, HT); chocolate chip ice cream (M, HT, Sus.); gold (PS, E); hand cream (M, HM, Sol’n); NaCl+H2O (M, HM, Sol’n); PowerAde (M, HM, Sol’n); Frappuccino (M, HM, Sol’n); Brass (M, HM, Sol’n); NaCO3 (PS, C); Vinagrette (M, HT, Sus.)
5. See pg 82-83, section 3.3: Filtration (to separate solids from liquids); Distillation (to separate two liquids with varying boiling point); Crystallization (to separate solute from solvent in a solution); Sublimation (to separate a mixture of solids when one of the solids can sublime); chromatography (to separate two substances in a solution)
6a. 70g
6b. the “3” grams was not lost, it was converted into a gas, which was not trapped and measured
6c. If the gas given off from the candle was trapped and its mass was measured, it would have a mass of 3g. Therefore, the mass before the reaction (mass of total reactants, 73g) is equal to the mass after the reaction (mass of total products 73g)
7. C= 42.86%, O= 57.86%
8. %O= 72%. The two compounds are not the same because they don’t contain the %Oxygen by mass, in accordance with the Law of Definite Proportions. (57% O by mass for cmpd 1; 72% O by mass for cmpd 2)
9. mass ratio for cmpd 1: 0.75gofC/gofO; mass ratio for cmpd 2: 0.37gofC/gofO; The two compounds are different. The whole number ratio of cmpd1:cmpd2 in accordance with Law of Multiple Proportions is 2: cmpd 1 has 2x more gofC/gofO than cmpd 2 (0.75gofC/GofO per 0.37gofC/GofO= 2)
Do you need more practice problems? Can you just not get enough!!??!! See problems 71, 73-80, and 93 at the end of chapter 3. Answers posted below!!! Your welcome 🙂
71. LoDP
73. NaCl= 1:1, CuO= 1:1, H2O= 2:1, H2O2= 1:1
74. 3.16% –> 3% w. SF
75. 39.7% (3SF)
76. 92.96% –> 93% w. 3SF
77. cmpd1: 0.75gC/gO; cmpd2: 0.37gC/gO; ratio of cmpd1:cmpd2: whole number = 2
78. 64%
79. Law of Multiple Proportions. CO2= higher percent by mass of O because more O atoms
80. CuO: 19%, 68.0g; H2O: 89%, 1.98g; H2O2: 94%, 32.0g; CO: 57%, 12.0g; CO2: 73%, 11.9g
93a. Samples I, IV, and III are from the same compound bc they have the same mass ratio of y:x (as shown by the same slope)
93b. approximate ratio: ~0.25gY/gX
93c. approximate ratio: ~0.5gY/gX
Matter Concept Maps
