Mole and Particle practice
How many representative particles (atoms, ions, molecules, etc) are present in each of the following?
1) 4.3 mol of tungsten
2) 2.45 * 10-6 mol nickel (II) selenide (NiSe)
3) 0.923 mol of selenium tetrabromide (SeBr4)
How many moles are equivalent to each of the following?
4) 7.95 * 1024 copper (II) chloride
5) 6.93 * 1023 thallium atoms
6) 1.974 * 1014 sodium chloride formula
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How many representative particles (atoms, ions, molecules, etc) are present in each of the following?
1)4.3 mol of tungsten
2)2.45 * 10-6 mol nickel (II) selenide (NiSe)
3)0.923 mol of selenium tetrabromide (SeBr4)
There are 6.02 x 10^23 molecules in a mole of molecules or 6.02 x 10^23 atoms in a mole of atoms.
4.3 moles W x (6.02 x 10^23 atoms W/1 mole W) = ?? atoms W. Note how the unit you don't want (moles W in this case) cancel and leaves the unit you want to keep (in this case atoms W).
I still don't understand can you help me with the first two?
Think of it this way. There are 40 apples in a bushel. How many apples in 3 bushels?
3 bushels x (40 apples/bushel) = 120 apples. See the unit bushel cancel leaving the unit we want to keep as apples.
4.5 x 10^-6 mol NiSe. There are 6.02 x 10^23 molecules in a mol, so
4.5 x 10^-6 mol NiSe x (6.02 x 10^23 molecules NiSe/mol NiSe) x (2 particles in NiSe/molecule NiSe) = number of particles in 4.5 x 10^-6 moles NiSe.
Again, note mol NiSe in the first term (4.5 x 10^-6 mol NiSe) cancel with mol NiSe in the denominator of the first factor and molecules NiSe in the numerator of the first factor cancel with molecule NiSe in the denominator of the second factor. We wanted to get rid of those units and we did. What's left? The unit we want to keep is particles and that's the only unit that didn't cancel.
To determine the number of representative particles (atoms, ions, molecules, etc) present in a given number of moles, we use Avogadro's number, which states that 1 mole of any substance contains 6.022 x 10^23 representative particles.
Let's solve each problem step by step:
1) 4.3 mol of tungsten:
To find the number of representative particles, we multiply the number of moles by Avogadro's number:
4.3 mol × 6.022 x 10^23 representative particles/mol = 2.60 x 10^24 representative particles
Therefore, there are 2.60 x 10^24 representative particles in 4.3 mol of tungsten.
2) 2.45 * 10^-6 mol nickel (II) selenide (NiSe):
Similarly, we multiply the number of moles by Avogadro's number:
2.45 x 10^-6 mol × 6.022 x 10^23 representative particles/mol ≈ 1.47 x 10^18 representative particles
Therefore, there are approximately 1.47 x 10^18 representative particles in 2.45 x 10^-6 mol of nickel (II) selenide.
3) 0.923 mol of selenium tetrabromide (SeBr4):
Again, we use the same approach:
0.923 mol × 6.022 x 10^23 representative particles/mol ≈ 5.55 x 10^23 representative particles
Therefore, there are approximately 5.55 x 10^23 representative particles in 0.923 mol of selenium tetrabromide.
Now, let's move on to the second set of problems:
4) 7.95 x 10^24 copper (II) chloride:
To determine the number of moles, we divide the given number of representative particles by Avogadro's number:
7.95 x 10^24 representative particles ÷ 6.022 x 10^23 representative particles/mol ≈ 13.2 mol
Therefore, 7.95 x 10^24 copper (II) chloride is equivalent to approximately 13.2 mol.
5) 6.93 x 10^23 thallium atoms:
Using the same approach:
6.93 x 10^23 representative particles ÷ 6.022 x 10^23 representative particles/mol ≈ 1.15 mol
Therefore, 6.93 x 10^23 thallium atoms are approximately equivalent to 1.15 mol.
6) 1.974 x 10^14 sodium chloride formula units:
Again, we apply the same principle:
1.974 x 10^14 representative particles ÷ 6.022 x 10^23 representative particles/mol ≈ 0.0328 mol
Therefore, 1.974 x 10^14 sodium chloride formula units are approximately equivalent to 0.0328 mol.
To summarize:
1) 4.3 mol of tungsten contains 2.60 x 10^24 representative particles.
2) 2.45 x 10^-6 mol of nickel (II) selenide contains approximately 1.47 x 10^18 representative particles.
3) 0.923 mol of selenium tetrabromide contains approximately 5.55 x 10^23 representative particles.
4) 7.95 x 10^24 copper (II) chloride is approximately equivalent to 13.2 mol.
5) 6.93 x 10^23 thallium atoms are approximately equivalent to 1.15 mol.
6) 1.974 x 10^14 sodium chloride formula units are approximately equivalent to 0.0328 mol.