The teacher gave her students an unknown sample that had density of 0.75g/cm3. The students were instructed tobreak thesample in
to 3 pieces. How did breaking the sample into 3 pieces impact their density
2 answers:
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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 mass of PbSO4 formed 15.163 gram
Explanation:
mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625
mole of Na₂SO₄ = 2 x 0.025 = 0.05
Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃
( Mole/Stoichiometry )
= 0.0625 = 0.05
From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.
Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄
= 0.05 x 303.26 g
= 15.163 g
There are:
3.41 moles of C
4.54 moles of H
3.40 moles of O.
Why?
To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.
Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.
We have that:
To know the percent of each element, we need to to the following:
So, we know that for the 100 grams of the compound, there are:
40.92 grams of C
4.58 grams of H
54.49 grams of O
We know the molecular masses of each element:
Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:
Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.
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