C is the answer I took a test with the same answer
Answer:
MoClBr₂
Explanation:
First we calculate the mass of bromine in the compound:
- 300.00 g - (82.46224 g + 45.741 g) = 171.79676 g
Then we<u> calculate the number of moles of each element</u>, using their <em>respective molar masses</em>:
- 82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo
- 45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl
- 171.79676 g Br ÷79.9 g/mol = 2.150 mol Br
Now we <u>divide those numbers of moles by the lowest number among them</u>:
- 0.9594 mol Mo / 0.9594 = 1
- 1.290 mol Cl / 0.9594 = 1.34 ≅ 1
- 2.150 mol Br / 0.9594 = 2.24 ≅ 2
Meaning the empirical formula is MoClBr₂.
Answer:
3
three half-filled orbitals each capable of forming a single covalent Bond and an additional lone - pair of electrons
Answer:
Molarity= 1.69M
Explanation:
m= 14.8, Mm= 35, V= 0.25dm3, C= ?
Moles = m/M= C×V
Substitute and Simplify
m/M= C×V
14.8/35= C×0.25
C= 1.69M
