Answer:
83.8%
Explanation:
The balanced reaction equation is;
2Al(s) + 3Cl2(g) → 2AlCl3(s)
Now we have to obtain the limiting reactant as the reactant that produces the least amount of AlCl3
Amount of Al = 3.11g/27 g/mol = 0.115 moles
If 2 moles of Al yields 2 moles of AlCl3
Then 0.115 moles of Al yields 0.115 moles of AlCl3
For Cl2
Amount of Cl2 = 5.32 g/71 g/mol= 0.075 moles
If 3 moles of Cl2 yields 2 moles of AlCl3
0.075 moles of Cl2 yields 0.075 * 2/3 = 0.05 moles of AlCl3
Hence Cl2 is the limiting reactant
Theoretical yield of AlCl3 = 0.05 moles of AlCl3 * 133g/mol = 6.65 g
%yield = actual yield /theoretical yield * 100
%yield = 5.57 g/6.65 g * 100
%yield = 83.8%
Answer:
7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.
Explanation:
First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.
According to the given question we have to prepare 0.100 M solution
1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4
1 ml of solution contain 14.208÷1000= 0.014 gram
0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.
So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.
Answer:
the spiral type of galaxy
Answer:
The classification and illustrations are attached in the drawing.
Explanation:
It is possible to identify the pure substance observing the figure, since it is the only one that has 2 joined atoms (purple and blue) which forms a single compound.
On the other hand, the homogeneous mixture is identified by noting that its atoms are more united with respect to the heterogeneous mixture, highlighting that in homogenous mixtures the atoms, elements or substances are not visible to the naked eye and are in a single phase, instead in the heterogeneous mixture if they can be differentiated.
Answer:
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