Apiece of aluminum foil 1.00cm square and 0.590mm thick is allowed to react with bromine to form aluminum bromide. part a how many moles of aluminum were used? (the density of aluminum is 2.699 g/cm^3 part b how many grams of aluminum bromide form, assuming that the aluminum reacts completely?

Part A: 5.899x10^-3 moles of Al

Part B: 1.573 g of AlBr3

Explanation:

Part A: We have to obtain the volume of the piece of aluminium; all sides of the square must be in cm. Then, use the density to obtain the mass.

0.059 is the volume of the Al udes for the reaction. Now, to oabtain the moles:

Part B: To obatin the mass of AlBr3, we need the balanced chemical equation:

                                               2Al + 3Br2 → 2AlBr3

We assume bromine (Br2) is in excess, therefore, we calculate the aluminum bromide formed from the Al:

of Al

the answer is a. co2, this would be colder than the : )

explanation:

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).