The question in English is "<span>Determine the mass, in kg, of a material that is contained in a volume of 18L. It is known that the material density is 0.9 g/cm 3"
Answer:
</span>
We can use a simple
equation to solve this problem. <span>
d =
m/v</span><span>
<span>Where </span>d <span>is
the density, </span>m <span>is
the mass and </span>v is the volume.
d = </span>0.9<span> g/cm³
m = ?
v = </span>18 L = 18 x 10³ cm³<span>
By applying the equation,
<span> 0.9 g/cm³ = m / </span></span>18 x 10³ cm³<span>
m = 0.9 g/cm³ x </span>18 x 10³ cm³<span>
<span>
</span>m = 16200 g
m = 16.2 kg
Hence, the mass of
18 L of material is 16.2 kg.</span>
1. The balanced equation tells us that 2 moles of H2S produce 2 moles of H2O.
8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced
8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically
% yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8 % yield
2. Calculate moles of each reactant.
150.0 g N2 x (1 mole N2 / 28.0 g N2) = 5.36 moles N2
32.1 g H2 x (1 mole H2 / 2.02 g H2) = 15.9 moles H2
The balanced equation tells us that we need 3 moles of H2 to react with every 1 mole of N2.
So if we have 5.36 moles N2, we need 3x that = 16.1 moles H2. Do we have that much available? No, just under at 15.9 moles. So H2 is the limiting reactant. At the end of the reaction there will be a little N2 left over.
<span>
</span>The solution would be like this for this
specific problem:
<span> (atomic number) - (core electrons)</span>
Boron:
Atomic number: 5
Core electrons: 2
Boron will have the
following: 2 core electrons and 3 valence electrons.
(atomic number) - (core electrons)------> (5) - (2) = +3
<span>
</span><span>Oxygen: </span><span>
<span>Atomic number: 8 </span>
<span>Core electrons: 2 </span></span>
<span>Boron will have the
following: 2 core electrons and 6 valence electrons. </span><span>
<span>(atomic number) - (core electrons)------> (8) - (2) = +6 |
So the answer would be +3 and +6</span>.</span><span>
I hope this helps and if you have any further questions, please don’t hesitate
to ask again. </span>
