Part A Name the complex ion [Fe(CN)6]^3- . The oxidation number of iron is +3. Part B Name the complex ion [Cu(NH3)2(H2O)4]^2+ .
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Answer:
2 water + sugar + lemon juice → 4 lemonade
Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=
236.59g/mol
946.36g
=4mol
Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=
225g/mol
196.86g
=0.8749mol
Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=
257.83g/mol
193.37g
=0.7499mol
Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=
719.42g/mol
2050.25g
=2.8498mol
As we can see that number of moles of lemon juice are limited.
So, we will consider the reaction will complete in accordance with moles of lemon juice.
1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:
\frac{2}{1}\times 0.7499 mol = 1.4998 mol
1
2
×0.7499mol=1.4998mol of water
Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g
Water remained unused = 946.36 g - 354.8376 g =591.5223 g
1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:
\frac{1}{1}\times 0.7499 mol = 0.7499 mol
1
1
×0.7499mol=0.7499mol of water
Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g
Sugar remained unused = 196.86 g - 28.1325 g
1 mole of lemon juice gives 4 moles of lemonade.
Then 0.7499 mol of lemon juice will give:
\frac{4}{1}\times 0.7499 mol=2.996 mol
1
4
×0.7499mol=2.996mol of lemonade
Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g
Theoretical yield of lemonade = 2157.9722 g
Experimental yield of lemonade = 2050.25 g
Percentage yield of lemonade:
\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100
theoretical yield
Experimental yield
×100
\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%
2157.9722g
2050.25g
×100=95.00%
Answer:
Normality N = 0.2 N
Explanation:
Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.
Normality is represented by N.
Mathematically, we have :
Given that:
number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent
volume of solution (HCl) = 450 mL 450 × 10⁻³ L
Normality N = 0.2 N