Answer:
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Explanation:
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
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Answer:
The name of this compound is :
Bi2(CO3)3 = Bismuth Carbonate
Explanation:
The name of the compound is derived from the name of the elements present in it.
The rule followed while naming the compound are:
1. The first element (always the cation) is named as such .
2. The second element (The anion) end with "-ate , -ide ," etc
3. NO prefix is added while naming the first element.
For example : Bi2 can't be named as Dibismuth
Na2 = Can't be named as disodium
Hence the compound :
Bi2(CO3)3 contain two element : Bi and CO3. Here , Bi = cation (named as such) and CO3 = anion (named according to rules)
Bi = Bismuth
CO3 = carbonate
Bi2(CO3)3 = Bismuth Carbonate
The molecular mass of this compound is :
Molecular mass = 2 (mass of Bi) + 3(mass of C) + 6(mass of O)
= 2 (208.98)+3(12.01)+6(15.99)
= 597.987 u
Answer:
345.44 g or 0.34544 kg
Explanation:
Applying
D = m/V...................... Equation 1
Where D = Density of mercury, m = mass of mercury, V = Volume of mercury.
make m the subject of the equation
m = D×V................. Equation 2
From the question,
Given: D = 13.6 g/mL = 13.6 g/cm³, V = 25.4 cm³
Substitute these values into equation 2
m = 13.6×25.4
m = 345.44 g
m = 0.34544 kg
Hence the mass of mercury is 345.44 g or 0.34544 kg
