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.

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.

where is the graph? need more info