The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
The answer is potassium magnate
When HCl is added to metal ions, metal chlorides are produced. In this problem, it is asked whether the given ions precipitate or not when added to HCl. According to the rule, all chlorides except Ag+, Pb 2+, Hg2 2+ are soluble. Hence the ion that would precipitate is only lead (II) ion.
Acids change the rate Corrosion and increase its temperature.
The grams of carbon dioxide that are in 35.6 liters of Co2 is calculates as below
calculate the number of moles of CO2
At STP 1 mole = 22.4 L
what about 35.6 liters
= 1mole x 35.6 liters/ 22.4 liters = 1.589 moles
mass of CO2 = moles x molar mass of CO2
= 1.589 mol x 44 g/mol = 69.92 grams
