Water is an example of a compound.
a. true
b. false

Skills of homo sapiens were found an excavation. the skulls were preserved because the bodies were frozen. so, these fossils are (blank) fossils.the image shows the evolution of skulls beginning 2 to 3 million years ago. based on the image, modern human skulls(blank) ape skulls.

The last few miles of the marathon are the most difficult for heather, her hair plastered to her head, sweat clinging to her arms, and her legs already feeling as if they had nothing left, just dead weight. after grabbing a cup of ice water, she feels the ice cubes smash against her nose as she gulps some cool refreshment and keeps on running. in these last few miles, the breeze kicks up and she finally feels some coolness against her skin. drips of sweat, once clinging to her forehead, now spill down, and heather feels more pain as the sweat flows into her eyes.which of the following is the most likely reason why the ice struck heather’s nose when she took a drink? a) water can function as a solvent. b) water can store large amounts of heat. c) water can moderate temperatures through evaporative cooling. d) the density of water decreases when it freezes. e) water has a cohesive nature.sweat remained on heather’s forehead and arms because of the a) high salt content of sweat b) cohesive nature of water c) ability of water to moderate heat d) high evaporative cooling effect of water e) ability of water to act as a solvent

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes