A sample of phosphonitrilic bromide, PNBr2, contains 2.01 mol of the compound. Determine the amount (in mol) of each element present and the number of atoms of each element present in the sample.

2.01 moles of P → 1.21×10²⁴ atoms

2.01 moles of N → 1.21×10²⁴ atoms

4.02 moles of Br → 2.42×10²⁴ atoms

Explanation:

We begin from this relation:

1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br

Then 2.01 moles of PNBr₂ will have:

2.01 moles of P

2.01 moles of N

4.02 moles of Br

To determine the number of atoms, we use the relation:

1 mol has NA (6.02×10²³) atoms

Then: 2.01 moles of P will have (2.01  . NA) = 1.21×10²⁴ atoms

2.01 moles of N (2.01  . NA) = 1.21×10²⁴ atoms

4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms

For 2.01 moles PNBr2 we have 1.21 *10^24 molecules

P =  1.21 *10^24 atoms

N = 1.21 *10^24 atoms

Br = 2.42*10^24 atoms

Explanation:

Step 1: Data given

Number of moles of PNBr2 = 2.01 moles

Step 2: Calculate moles per element

For 1 mol PNBr2 we have 1 mol P, 1 mol N and 2 moles Br

For 2.01 moles PNBr2 we have:

P = 2.01 moles

N = 2.01 moles

Br = 2* 2.01 = 4.02 moles

Step 3: Calculate atoms of each element

For 2.01 moles PNBr2 we have 6.022 *10^23 * 2.01 = 1.21 *10^24 molecules

P = 2.01 moles = 6.022 *10^23 * 2.01 = 1.21 *10^24 atoms

N = 2.01 moles = 6.022 *10^23 * 2.01 = 1.21 *10^24 atoms

Br = 4.02 moles = 6.022 *10^23 * 4.02 = 2.42*10^24 atoms

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.