The first step in the proton-proton chain produces an antielectron, or positron. What happens to the positron?
A. It slowly works its way to the Sun's surface, where it escapes into space.
B. It rapidly escapes from the Sun, traveling into space at nearly the speed of light.
C. It is rapidly converted to energy when it meets an ordinary electron, resulting in matter antimatter annihilation
OD. It quickly meets an ordinary electron, forming an electron-positron pair that remains stable.
E. It joins with a nearby neutron to form a proton.

Given the following at 25c calculate delta hf for hcn (g) at 25c. 2nh3 (g) +3o2 (g) + 2ch4 (g) > 2hcn (g) + 6h2o (g) delta h rxn= -870.8 kj. delta hf=-80.3 kj/mol for nh3 (g), -74.6 kj/mol for ch4, and -241.8 kj/mol for h2o (g)

Which of these is an intensive property of a substance

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