SAMPLE A - <span>pure substance.
</span>SAMPLE B - <span>homogeneous mixture.
</span>SAMPLE C - <span>heterogeneous mixture.
</span>Pure substance - <span>constant composition and properties.</span>
Homogeneous mixture - same uniform appearance and composition.
Heterogeneous mixture - <span>not </span>uniform<span> in composition, two phases (liquid and dust).
</span>
The answer is B. Unit cell.
Hoped I Helped!
The hydrate form of CuSO4 has 5 water molecules (CuSO4-5H20) copper (II) Sulfate pentahydrate or commonly known as blue vitriol.
To solve, the following molar masses are to be known.
CuSO4.5H2O (hydrate) - 249.7g/mole
CuSO4 (anhydrous) -159.6g/mole
Also there molar ratio of the hydrate and CuSO4 is 1.
the mass of the hydrate is to be divided by the molar mass of the hydrate then multiplied by the ratio (1) to get the moles of hydrate and multiplied by the molar mass of the anhydrous to get the mass in grams.
moles = (100g/249.7)*1 = 0.4 moles hydrate
grams = 0.4*159.6 = 64.9 grames hydrate
Concentration of Ba2+ is 1.00 mol/dm3 or 1M
<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>
