<em>Answer:</em>
The order of liquid from top to bottom is as follow
<em>Explanation:</em>
Chart of densities:
- Red = 1.2 g/cm∧3
- Blue = 1.6 g/cm∧3
- green = 0.8 g/cm∧3
- Purple = 0.1 g/cm∧3
<em>Density: </em>
Density is the ratio of mass and volume as follow
d = m/v
<em>Summary:</em>
- Greater the density, it will be at bottom and vice versa.
- Blue liquid has greater density so it will be at bottom
- Purple liquid has low density, so it will be at top.
Answer:
C3H6O2
Explanation:
To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.
H = 4.10/1.36 = 3
O = 1.36/1.36 = 1
C = 2.05/1.36 = 1.5
We then multiply through by 2 to yield the compound with the empirical formula C3H6O2
Answer:
The number of moles of potassium hydroxide, KOH required to make 4 moles of K₂SO₄ is 8 moles of KOH
Explanation:
2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
From the above reaction, we have 2 moles of KOH combining with 1 mole of H₂SO₄ to produce 1 mole of K₂SO₄ and 2 moles of H₂O.
Therefore the number of moles of potassium hydroxide that will be needed to make 4 moles of K₂SO₄ is;
8KOH + 4H₂SO₄ → 4K₂SO₄ + 8H₂O
8 moles of KOH is required to make 4 moles of K₂SO₄.
In a chemical reaction,
the limiting reagent is the chemical being used up while the excess reactant is
the chemical left after the reaction process.
Before calculating the limiting
and excess reactant, it is important to balance the equation first by stoichiometry.
C25N3H30Cl + NaOH = C25N3H30OH + NaCl
Since the reaction is already balanced, we can now identify which
is the limiting and excess reagent.
First, we need to determine the number of moles of each chemical
in the equation. This is crucial for determining the limiting and excess reagent.
<span>Assuming that there is the
same amount of solution X for each reactant</span>
1.0 M NaOH ( X ) = 1.0
moles NaOH
1.00 x 10-5 M C25N3H30Cl
( X ) = 1.00 x 10-5 moles C25N3H30Cl
<span>The result showed that the
crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this
chemical reaction is crystal violet and the excess reactant is NaOH.</span>
