Answer:
1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.
2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.
Explanation:
The molecular formula of boron trifluoride is
.
So, one mole of boron trifluoride has one mole of boron atoms.
1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.
The number of atoms in 2.20 moles of boron is:
One mole of boron has ----
atoms.
Then, 2.20 moles of boron has
-
2. Calculate the number of moles of BF3 in 5.35*1022 molecules.

One mole of boron trifluoride has three moles of fluorine atoms.
Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.
=0.266mol of fluorine atoms.
Answer:
The following formula is used to calculate the speed or velocity of a wave. V = f * w Where V is the velocity (m/s) f is the frequency (hz)
Answer:
Two hundred million years ago, Antarctica was a lush, temperate rainforest home to crocodile-sized amphibians and rhinoceros-sized dinosaurs.
Explanation:
Animals:
1 ankylosaurs (the armored dinosaurs), mosasaurs and plesiosaurs (both marine reptilian groups).
2 Cry-oloph-os-auru-s was the ap-ex pre-da-tor of its region when it hunted in the lands of Jur-as-sic Period Antarctica.
3 plesiosaurs.
4 mosasaurs.
Plants:
1 palm trees.
2 ferns and Tanglefoot.
3 Evergreen trees
4 Glossopteris
Answer:
Concentration of unknown solution is 0.0416 M
Explanation:
As we know
Absorbance is equal to the product of molar absorptivity of KMnO4 m, path length and concentration
From the given set of graphical data, it is clear that the absorbance vs concentration is a straight line.
From the graph, we can obtain-
Y = 5.73 X – 0.0065
Absorbance = 0.232
0.232 = 5.73 X – 0.0065
X = 0.0416
Concentration of unknown solution is 0.0416 M
Answer:
The answer to your question is given after the questions so I just explain how to get it.
Explanation:
a)
Get the molecular weight of Phosphoric acid
H₃PO₄ = (3 x 1) + (31 x 1) + (16 x 4)
= 3 + 31 + 64
= 98 g
98 g ----------------- 1 mol
0.045 g --------------- x
x = (0.045 x 1) / 98
x = 0.045 / 98
x = 0.00046 moles or 4.6 x 10 ⁻⁴
b)
Molarity = 
Molarity = 
Molarity = 0.0013 or 1.31 x 10⁻³
c)
Formula C₁V₁ = C₂V₂
V₁ = C₂V₂ / C₁
Substitution
V₁ = (0.0013)(1) / 0.01
Simplification and result
V₁ = 0.0013 / 0.1
V₁ = 0.13 l = 130 ml