Answer:
0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).
Explanation:
<em>d = m/V,</em>
where, d is the density of the material (g/cm³).
m is the mass of the material (m = 28 g).
V is the volume of the material (V = 63.0 cm³).
<em>∴ d = m/V </em>= (28 g)/(63.0 cm³) = <em>0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).</em>
Answer:
A suction bulb or pipette pump to suction the sample.
Explanation:
There are different types of pipettes that can be used in a laboratory:
- Volumetric pipette-used to transfer a specific calculated amount of liquid
- Graduated pipettes-used to transfer different calculated amounts of liquid from one container to another.
- Micropipettes-used to transfer small amounts of liquids
- Pasteur pipettes- just like micropipettes, pasteur pipettes are used to transfer small amounts of liquids, but manually. They are also called droppers.
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.
An isotope is different from an element by the number of neutrons it has.