Initially, a particular sample has a total mass of 200 grams and contains 128 x 1010 radioactive nuclei. these radioactive nuclei have a half life of 1 hour. (a) after 3 hours, how many of these radioactive nuclei remain in the sample (that is, how many have not yet experienced a radioactive decay)

Explanation:

Formula to calculate how many particles are left is as follows.

              N =

where,     = number of initial particles

                              l = number of half lives

As it is given that number of initial particles is and number of half-lives is 3.

Hence, putting the given values into the above equation as follows.

               N =

                    =

                    =

or,                    

Thus, we can conclude that particles of radioactive nuclei remain in the given sample.

In five hours we've gone through 5 half lives so the answer is:

particles

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72