1. how many grams of b are present in 3.35 grams of boron tribromide ?
grams b.
2. how many grams of boron tribromide contain 4.69 grams of br ?
boron tribromide.
3. how many grams of n are present in 4.11 grams of nitrogen trifluoride ?
n.
4. how many grams of nitrogen trifluoride contain 3.07 grams of f ?
nitrogen trifluoride.
5.how many grams of co3+ are present in 1.16 grams of cobalt(iii) iodide?
co3+.
6. how many grams of cobalt(iii) iodide contain 2.28 grams of co3+?
cobalt(iii) iodide.

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

grams B.

MW BBr₃ = 251 g

                            251 g of BBr₃  11 g of B

                             3.35 g              x

                           x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

grams boron tribromide.

MW BBr₃ = 251g

                             251g of BBr₃   80 g of Br

                                   x               4.69 g

                           x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

grams N.

MW NF₃ = 71 g

                               71 g of NF₃       14 g of N

                               4.11 g                   x

                               x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

grams nitrogen trifluoride.

MW NF₃ = 71

                                71 g of NF₃   19 g of F

                                  x                 3.07 g

                                 x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

grams Co3+.

MW CoI₃ = 440 g

                             440 g of CoI₃  59 g of Co

                              1.16 g               x

                              x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

grams cobalt(III) iodide.

MW CoI₃ = 440 g

                             440 g of CoI₃  59 g of Co

                               x                     2.28 g of Co⁺³

                              x = (2.28 x 440) / 59

                              x = 17 g of CoI₃

the answer is definitely c.

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

: )