A. The concentration ratio between trials 2 and 1 is: 0.3 / 0.1 = 3.
B. The ratio between reaction rates is: 0.054 / 0.006 = 9.
C. We solve the equation: (concentration ratio)^x = (reaction rate ratio):

, which gives x = 2, and the exponent in the rate law is 2.
D. The rate law will have the equation: r = k[NO2]^2
E. To solve for k, we can substitute the values for [NO2] = 0.1, rate = 0.006. This gives:

, which yields k = 0.6 L/mol-s.
F. Since the exponent is 2, this is a second-order reaction.
Answer:
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Explanation:
Answer:
A) oxidizing agent is SO2
B) NaClO is the oxidizing agent
Explanation:
A) This is a redox reaction in which oxidation and reduction occur simultaneously.
Thus, in 2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s);
H2S is reduced as follows;
H2S → S + 2H+ + 2e−
We can see that SO2 has been reduced while H2S gets oxidized since it has changed state from - 2 to 0 . Thus sulphur dioxide is the oxidizing agent.
B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)
In this, SO2 undergoes oxidation and NaClO is the oxidizing agent
Answer:
Because energy is conserved, the kinetic energy of a block at the bottom of a frictionless ramp is equal to the gravitational potential energy of the block at the top of the ramp. This value is proportional to the square of the block’s velocity at the bottom of the ramp. Therefore, the block’s final velocity depends on the height of the ramp but not the steepness
of the ramp
Explanation: