Answer:
They are different ions of the same element.
Explanation:
on Quizlet
Theoretical
yield is the amount supposedly produced by the reaction if it was complete. The
balanced chemical reaction is:<span>
</span>
<span> C2H6 + Cl2 = C2H5Cl + HCl</span>
<span>
We use the amounts given for the reactants as
the starting point of the calculations to determine the limiting reactant. We do as follows:
150 g C2H6 ( 1 mol / 30.08 g
) = 4.987 mol
205 g Cl2 ( 1 mol / 70.9 g ) = 2.891 mol
Therefore, the limiting reactant would be Cl2 since it would be consumed completely first in the reaction.
We use this for the theoretical yield.
<span>2.891 mol Cl2 ( 1 mol C2H5Cl / 1 mol Cl2 ) = </span>2.891 mol C2H5Cl (THEORETICAL YIELD)
172 g ( 1 mol / 64.52 g ) = 2.666 mol C2H5Cl (
ACTUAL YIELD)
Percent yield = 2.666 / 2.891 x 100 = 92.21%</span>
The answer is repeated dehydration synthesis reactions.
That is the formation of a complex carbohydrate is a type of repeated dehydration synthesis reactions.
In dehydration synthesis reactions, two or molecules of sugar are joined together by the removal of water molecule that is why it is known as dehydration synthesis reaction.
In carbohydrates these reactions results in the formation of polysaccharide.
Since it talks about the number of particles, we would need the Avogadro's number which is an empirical value equal to 6.022×10²³ particles/mole. The other information that we have to know is the molar mass of Na₂SO₄ which is 142.04 g/mol.
Mass of a single particle of Na₂SO₄:
(142.04 grams Na₂SO₄/mol) * (1 mol/6.022×10²³ particles) = 2.359×10⁻²² g
Thus, each Na₂SO₄ weighs 2.359×10⁻²² g.
Number of Na₂SO₄ particles in the mixture:
(11.53 g Na₂SO₄) * (1 mol/142.04 g) * (6.022×10²³ particles/mol) = 4.89×10²² particles
Thus, the mixture contains 4.89×10²² particles is Na₂SO₄.
Answer:
1) The power needed to process 50 ton/hr is 135.4 HP.
2) The void fraction of the bed is 0.37.
Explanation:
1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).
We assume the units of Ei are kWh/t.
The equation that relates this parameters and the power is (size of particles in μm):
The power needed to process 50 ton/hor is
2) The density of the packed bed can be expressed as
being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).
Then we can rearrange
The void fraction of the bed is 0.37.