Aluminium belongs to 13th group of periodic table. It undergoes oxidation to given Al^+3 .
It is observed that when aluminium is added to a solution of copper sulphate the colour of the solution changes from blue to grey. It is due to formation of grey coloured solution of aluminium sulphate as
2Al^+3 + 3SO4^-2 ---> Al2(SO4)3
Answer:
a) The theoretical yield is 408.45g of 
b) Percent yield = 
Explanation:
1. First determine the numer of moles of 
and 
.
Molarity is expressed as:
M=
- For the
M=
Therefore there are 1.75 moles of
- For the
M=![\frac{2.0moles[tex]Na_{2}SO_{4}](https://tex.z-dn.net/?f=%5Cfrac%7B2.0moles%5Btex%5DNa_%7B2%7DSO_%7B4%7D)
}{1Lsolution}[/tex]
Therefore there are 2.0 moles of
2. Write the balanced chemical equation for the synthesis of the barium white pigment, :
3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the :
- For the :
As the is the smalles quantity, this is the limiting reagent.
4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.
5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield = 
Percent yield =
The real yield is the quantity of barium white pigment you obtained in the laboratory.
Answer:
Iron has 5 unpaired electrons in Fe⁺³ state.
Explanation:
Iron having atomic number 26 has following electronic configuration in neutral state.
Fe = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁶
When Iron looses three electrons it attains +3 charge with following electronic configuration.
Fe⁺³ = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵
The five electrons in d-orbital exist in unpaired form as,
3(dz)¹, 3d(xz)¹, 3d(yz)¹, 3d(xy)¹, 3(dx²-y²)¹
Answer:
Option A.
2Na + 2H2O —> 2NaOH + H2
Explanation:
To know which option is correct, we shall do a head count of the number of atoms present on both side to see which of them is balanced. This is illustrated below below:
For Option A:
2Na + 2H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
4 H >>>>>>>>>>>> 4 H
2 O >>>>>>>>>>>> 2 O
Thus, the above equation is balanced.
For Option B:
2Na + 2H2O —> NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 3 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
For Option C:
2Na + H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
2 H >>>>>>>>>>>> 4 H
1 O >>>>>>>>>>>> 2 O
Thus, the above equation is not balanced.
For Option D:
Na + 2H2O —> NaOH + 2H2
Reactant >>>>>>> Product
1 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 5 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
From the illustrations made above, only option A is balanced.
