Answer:
for me to be able to get to the office today and I will be in the office tomorrow and will be able the get time final
194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
Explanation:
In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.
It is known that 1 moles of any element has 6.022×10²³ molecules.
Then 1 molecule will have
moles.
So 
Thus, 1.66 moles are included in BCl₃.
Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.
As it is known as 1 mole contains molecular mass of the compound.
As the molecular mass of BCl₃ will be

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.
Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.


So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
Answer:
+6
Explanation:
We are given;
We are required to determine the oxidation number of S in the compound.
We need to know that;
- The total oxidation number of this compound is Zero
- Oxidation number of Ba metal is +2
- Oxidation number of Oxygen atom is -2
- There is 1 atom of Ba, 1 atom of S and 4 atoms of O in the compound.
Therefore; assuming oxidation number of S is x
Then, (1× 2) + (1 × x) + ( 4 ×(-2)) = 0
2 + x - 8 = 0
x = 6 or +6
Thus oxidation number of Sulfur in BaSO₄ is +6.
Answer:
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Explanation:
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