Answer:
81.97 g of NaAl(OH)₄
Solution:
The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,
2 NaOH + 2 Al + 6 H₂O → 2 NaAl(OH)₄ + 3 H₂
According to this equation,
2 Moles of NaOH produces = 163.94 g (2 mole) of NaAl(OH)₄
So,
1 Mole of NaOH will produce = X g of NaAl(OH)₄
Solving for X,
X = (1 mol × 163.94 g) ÷ 2 mol
X = 81.97 g of NaAl(OH)₄
Answer:
I think it would be b. The octet rule states that transition metal group elements tend to react so that they attain a noble gas electron configuration.
It is lead (ll) iodide or just lead iodide
The melting point of any substance is its physical property.
colour of any substance is also its physical property.
However TNT is a chemical property. It is explosive as it dissociates into nitrogen gas , water, Carbon monoxide and carbon.
the reaction can be given as
2 C7H5N3O6 ---> 3 N2 + 5 H2O + 7 CO + 7 C
Due to carbon it gives as sooty flames
Neutralization is the name of the reaction
