T(s) [A](M) ln[A] 1/[A]0.00 0.500 −0.693 2.0020.0 0.389 −0.944 2.5740.0 0.303 −1.19 3.3060.0 0.236 −1.44 4.2480.0 0.184 −1.69 5.43Part AWhat is the order of this reaction?012SubmitHintsMy AnswersGive UpReview PartCorrectNow that we know that the reaction is first order, we know that the units of the rate constant will be inverse seconds, s−1.Part BWhat is the value of the rate constant for this reaction?Express your answer to three significant figures and include the appropriate units. k = 1.25×10−2 s−1SubmitHintsMy AnswersGive UpReview PartCorrectNext, the student ran the same reaction at a different temperature and measured a different reaction rate, with the following results:T(K) k(s−1)295 0.0125325 0.0357Part CWhat is the activation energy of this reaction?Express your answer to three significant figures and include the appropriate units. Ea =

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.