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According to its formula FeSO4.7H2O
we can get the percent % by the mass of H2O from this formula
%mass of H2O = (mass of water H2O/ mass of the hydrate)x100
when the mass of water = molar mass x 7 = 18 x 7 = 126
and the mass of hydrate (feSO4) = molar mass = 278
So by substitution:
%mass of H2O = (126/278) x 100 = 45%
Answer:
1. Iron(iii) hydroxide
2. It is a base
Explanation:
To determine the name of the compound, we must find the oxidation number Fe in the compound since the oxidation number of O is —2 in all its compound except for peroxide where it —1 and the oxidation number of Hydrogen is always +1 in all its compounds except for hydrides where it is —1. The oxidation number of Fe can calculated as follows:
Fe(OH)3 = 0
O = —2
H = +1
Fe =?
Fe + 3( —2 + 1) = 0
Fe + 3(—1) = 0
Fe —3 = 0
Fe = 3
The oxidation number of Fe in Fe(OH)3 is 3.
Therefore, the name of Fe(OH)3 is Iron(iii) hydroxide
Fe(OH)3 is a base since it contains the OH group
Answer:
<u>ATGGCCTA</u>
Explanation:
For this we have to keep in mind that we have a <u>specific relationship between the nitrogen bases</u>:
-) <u>When we have a T (thymine) we will have a bond with A (adenine) and viceversa</u>.
-) <u>When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa</u>.
Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:
TACCGGAT
<u>ATGGCCTA</u>
<u></u>
I hope it helps!
Answer:
The theoretical yield of Cu(s) in moles is 60.15 moles
Explanation:
Step 1: Data given
Number of moles CuO = 70.8 moles
Number of moles NH3 = 40.1 moles
Molar mass CuO = 79.545 g/mol
Molar mass NH3 = 17.03 g/mol
Step 2: The balanced equation
3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)
For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2
NH3 is the limiting reactant. It will completely be consumed (40.1 moles). CuO is in excess. There will react 3/2 * 40.1 = 60.15 moles
There will remain 70.8 - 60.15 = 10.65 moles CuO
Step 3: Calculate moles Cu
For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2
For 40.1 moles NH3 we'll have 60.15 moles Cu
The theoretical yield of Cu(s) in moles is 60.15 moles
