Answer:
The answer to your question is:
Step-by-step 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