Identify the balanced chemical equation that represents a decomposition reaction. p4 + 3o2 ⟶ p2o3
2fe(oh)3 ⟶ 2feo3 + h2o
cuso4 ⟶ cuo + 2so3
2fe(oh)3 ⟶ fe2o3 + 3h2o

Answer:

a. energy is absorbed in the reaction.

explanation:

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.

The answer to your question really depends on three things: how much water you have, how cold it is to start out, and how cold the things around it are. water actually freezes when it gets to 32 degrees fahrenheit (0 degrees celsius), but the time it takes to get there may be different.  let's start with the first. if you take two glasses, and fill one with a tiny bit of water, and the other about halfway, then put them both in the freezer, the one with less water will freeze first (you can try this at home, but i recommend using plastic cups and not glass ones).  now let's move on to the second part. let's say you have two glasses, and you fill one with really cold water that has been in the refrigerator, and the other with really hot water from the sink. if you put both of them in the freezer, the one that started out colder will freeze first.  for the third part, let's imagine that you have two glasses with the same amount of water in them, and the water is at the same temperature. imagine putting one outside on a really really cold day in georgia, and having a friend in alaska put one outside on the same day. since it would be so much colder in alaska, the glass of water there would freeze before yours.  so, if you took a tiny bit of really cold water in a glass, and put it outside on a cold day in alaska, it would freeze a lot faster than a big glass of hot water outside on a cold day in georgia. 

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.