Draw a Lewis structure for each covalent molecule. a. CH3F b. N2H4

See attached picture.

Explanation:

Hello,

In this case, given the compounds we notice:

a. Since carbon has four valence electrons (group IVA), it has three hydrogen atoms around it as well as a fluorine atom which has seven valence electrons (group VIIA). Moreover, we notice both carbon and fluorine attaining the octet.

b. In this case, two nitrogen atoms which have five valence electrons each (groupVA) are bonded via a single bond and two hydrogen atoms are bonded to each nitrogen atom in order to attain the octet for nitrogen only.

Thus, structures are given on the attached picture.

Regards.

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

може ли да зареждате повече информация, моля