Question

1. write the formula for chromium (iii) chloride (0.5 points) *

enter your answer

1. write the formula for chromium (ii) nitride (0.5 points) *

enter your answer

1. what is the name of ni2s3 (0.5 points) *

enter your answer

1. what is the name of v3p4 (0.5 points) *

enter your answer

Explanation:

Question 9

Step1: Determine the ions

Chromium (III) has a charge of \(Cr^{3 +}\), chloride has a charge of \(Cl^{-}\).

Step2: Balance the charges

To balance the charges, we need 3 chloride ions for 1 chromium ion.

Question 10

Step1: Determine the ions

Chromium (II) has a charge of \(Cr^{2+}\), nitride has a charge of \(N^{3 -}\).

Step2: Balance the charges

Using the criss - cross method, the formula is \(Cr_{3}N_{2}\) (by taking the absolute values of the charges as sub - scripts: for \(Cr\), charge \(+ 2\), for \(N\), charge \(-3\), so \(Cr_{3}N_{2}\)).

Question 11

Step1: Identify the elements and their charges

\(Ni\) is nickel. In \(Ni_{2}S_{3}\), let the charge of \(Ni\) be \(x\). For sulfur (\(S\)) in a simple sulfide, the charge is \(-2\). Using the charge balance equation \(2x+3\times(- 2)=0\), we get \(2x = 6\), \(x = + 3\).

Step2: Name the compound

The name is Nickel (III) sulfide.

Question 12

Step1: Identify the elements and their charges

\(V\) is vanadium. Let the charge of \(V\) be \(y\). For phosphorus (\(P\)) in a simple phosphide, the charge is \(-3\). Using the charge balance equation \(3y + 4\times(-3)=0\), we get \(3y=12\), \(y = + 4\).

Step2: Name the compound

The name is Vanadium (IV) phosphide.

Answer:

1. \(CrCl_{3}\)
2. \(Cr_{3}N_{2}\)
3. Nickel (III) sulfide
4. Vanadium (IV) phosphide