Answer:
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Explanation:
Empirical formula of the compound is the simplest ratio of elements present in the compound.
Empirical formula of compounds of chlorine with oxygen is as follows:
Compounds in which oxidation state of Cl is +1
Compounds in which oxidation state of Cl is +3
Compounds in which oxidation state of Cl is +4
Compounds in which oxidation state of Cl is +6
Compounds in which oxidation state of Cl is +7
The mass of 1.72 mol of magnesium fluoride is 107 grams.
To determine the mass of 1.72 mol of magnesium fluoride, we first need the chemical formula of magnesium fluoride. Magnesium forms a +2 ion (Mg+2) and fluoride forms a -1 ion (F-1). Since all compounds formed from ions have to be electrically neutral, we need 2 fluoride ions and 1 magnesium ion. Therefore, the formula for magnesium fluoride is MgF2.
Now we need to determine the molar mass of the compound from the molar mass values from the periodic table. Let's use a table to calculate this molar mass.
Molar mass of MgF2
Element Molar Mass (g/mol) Quantity Total (g/mol)
Mg 24.31 1 24.31
F 19.00 2 38.00
Total molar mass of MgF2 = 24.31 g/mol + 38.00 g/mol = 62.31 g/mol
This is the mass of one mole of the substance. If we have 1.72 mols of it, we multiply 1.72 by 62.31.
1.72 mol (62.31 g/mol) = 107 grams
We rounded to 107 to keep the correct number of significant digits in our answer.
The reaction between boron sulfide and carbon is given as:
2B2S3 + 3C → 4B + 3CS2
As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.
Given data:
Mass of C = 2.1 * 10^ 4 g
Mass of B = 3.11*10^4 g
Mass of CS2 = 1.47*10^5
Mass of B2S3 = ?
Now based on the law of conservation of mass:
Mass of B2S3 + mass C = mass of B + mass of CS2
Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5
Mass of B2S3 = 15.7 * 10^4 g
Answer: Trade winds :)
Explanation: just took the test
