Which atomic sub-particle has no charge?

Consider the following multistep reaction: a b→ab(slow) a ab→a2b(fast)−−−−−−−−−−−−−−−−− 2a b→a2b(overall) based on this mechanism, determine the rate law for the overall reaction. express your answer in standard masteringchemistry format. for example, if the rate law is k[a]3[b]2 type k*[a]^3*[b]^2

The ozone molecule o3 has a permanent dipole moment of 1.8×10−30 cm. although the molecule is very slightly bent-which is why it has a dipole moment-it can be modeled as a uniform rod of length 2.5×10−10 m with the dipole moment perpendicular to the axis of the rod. suppose an ozone molecule is in a 8000 n/c uniform electric field. in equilibrium, the dipole moment is aligned with the electric field. but if the molecule is rotated by a small angle and released, it will oscillate back and forth in simple harmonic motion.what is the frequency f of oscillation?

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