We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.
150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo
1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo
Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.
Answer:
a) The lewis dot structure is shown in the image attached to this answer
b) The formal charge on each of the atoms is zero
c) bromine has an oxidation state of +5 while fluorine has an oxidation state of -1
d) 90 degrees
e) Square Pyramidal
f) polar bonds
g) polar molecule
Explanation:
The molecule BrF5 has a formal charge of zero. It exhibits an sp3d2 hybridization state with a square pyramidal geometry. The bond angle in the molecule is 90 degrees. It is a molecule of the type AX5E. The oxidation state of bromine is +5 while that of fluorine is -1.
The Br-F bonds are polar. The overall molecule is polar due to asymmetric charge distribution concentrating on the central atom since the molecule is square pyramidal.
Empirical formula is the simplest ratio of whole numbers of components in a compound
in 100 g of compound
C H O
mass 25.5 g 6.40 g 68.1 g
number of moles 25.5 g/12 g/mol 6.40 g/ 1 g/mol 68.1 g/ 16 g/mol
= 2.13 mol = 6.40 mol = 4.26 mol
divide by least number of moles
2.13/2.13 = 1 6.40/2.13 = 3.0 4.26/2.13 = 2.0
all rounded off
C - 1
H - 3
O - 2
empirical formula - CH₃O₂
The answer to this is to prevent dilution of the base, which may affect the results. I think that is right.