Answer:
Explanation:
Oxidation state of Cr in CrPO₄
As a general rule, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero.
The compound above is in its neutral state and we sum all the oxidation numbers and equate to zero:
The oxidation number of P is -2
O is -2
Let the oxidation number of Cr be x:
x + (-2) + 4(-2) = 0
x -2-8 = 0
x -10 = 0
x = +10
For Cr in Cr₃(PO₄)₂
Using the same rule:
2(x) + 2[-2 + 4(-2)] = 0
2x + 2(-2-8) = 0
2x -20 = 0
x = +10
C. Oxidized and reduced are the same.
Density = mass / volume
Density = (45g) / (9 [volume units])
Density = 5g / [volume unit]
There was no specification on the units of volume. However, whatever it may be, just replace the square brackets with the value and your units will be correct.
