A sheet of aluminum measures 55.0 cm by 23.3 cm. If the volume is 2.90 cm^3 what is the thickness of the foil? Express your answer with the appropriate units.

0.0023 cm

Explanation:

The following data were obtained from the question:

Length (L) = 55 cm

Breadth (B) = 23.3 cm

Volume (V) = 2.9 cm³

Thickness (T) =?

Thus, we can obtain the thickness of the aluminum foil as illustrated below:

V = L × B × T

2.9 = 55 × 23.3 × T

2.9 = 1281.5 × T

Divide both side by 1281.5

T = 2.9/1281.5

T = 0.0023 cm

Therefore, the thickness of the aluminum foil is 0.0023 cm

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !