Question

assignment 2 part 1 9 marks choose 9 of 10 (lesson 4a & 4b) how many atoms of each element occur in each of the following compounds? a) selenide ion b) cobalt (iii) phosphate

Explanation:

Step1: Analyze selenide ion

The selenide ion is $Se^{2 - }$. It consists of 1 selenium (Se) atom.

Step2: Write formula for cobalt(III) phosphate

Cobalt(III) has a charge of $Co^{3+}$ and phosphate is $PO_{4}^{3 - }$. The formula for cobalt(III) phosphate is $Co_{3}(PO_{4})_{2}$.

Step3: Count atoms in cobalt(III) phosphate

For cobalt (Co), there are 3 atoms. For phosphorus (P), since there are 2 phosphate groups and each phosphate group has 1 P atom, there are 2 P atoms. For oxygen (O), since each phosphate group has 4 O atoms and there are 2 phosphate groups, there are $4\times2 = 8$ O atoms.

Answer:

a) 1 selenium atom
b) 3 cobalt atoms, 2 phosphorus atoms, 8 oxygen atoms