part 1 (1 point)\nrus\nruthenium (ii) sulphide\npart 2 (1 point)\npdcl₂\npalladium(ii) chloride\npart 3 (1…

part 1 (1 point)\nrus\nruthenium (ii) sulphide\npart 2 (1 point)\npdcl₂\npalladium(ii) chloride\npart 3 (1 point)\nag₂o\nsilver (i) oxide\npart 4 (1 point)\nwo₃\npart 5 (1 point)\npto₂\nplatinum(iv) oxide\n1st attempt

part 1 (1 point)\nrus\nruthenium (ii) sulphide\npart 2 (1 point)\npdcl₂\npalladium(ii) chloride\npart 3 (1 point)\nag₂o\nsilver (i) oxide\npart 4 (1 point)\nwo₃\npart 5 (1 point)\npto₂\nplatinum(iv) oxide\n1st attempt

Answer

Explanation:

Step1: Determine tungsten's oxidation state

In $WO_3$, oxygen has an oxidation - state of - 2. Let the oxidation state of tungsten be $x$. Using the rule that the sum of oxidation states in a neutral compound is 0, we have $x+3\times(- 2)=0$.

Step2: Solve for tungsten's oxidation state

$x - 6 = 0$, so $x = +6$.

Answer:

Tungsten(VI) oxide