What is the empirical formula of an oxide of chromium that is 48 percent oxygen by mass?

CrO₃.

Step-by-step explanation:

We'll begin by calculating the percentage of chromium in the oxide. This can be obtained as follow:

Oxygen (O) = 48%

Chromium (Cr) =?

Oxide of chromium contains only chromium and oxygen. Therefore, the percentage of chromium in the oxide is given by:

Cr = 100 – percentage of oxygen

Cr = 100 – 48

Cr = 52%

Finally, we shall determine the empirical formula of the oxide as follow:

Chromium (Cr) = 52%

Oxygen (O) = 48%

Divide by their molar mass

Cr = 52/52 = 1

O = 48/16 = 3

Divide by the smallest

Cr = 1/1 = 1

O = 3/1 = 3

Therefore, the empirical formula of the oxide is CrO₃.

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )

graph b

step-by-step explanation:

one solution- graph a (clear intercepting point.)

no solution- graph b (they're parallel, there will never be a point of interception.)

infinite solutions- graph c (all points intercept.)