Answer: Silicon the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Silicon go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons.

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Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g sample of a

kvv77 [185]

Answer:

The mass percent of Al(OH)₃ is 15.3%

Explanation:

The reaction is:

Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:

NaOH + HCl = NaCl + H₂O

The total moles of HCl is:

$n_{HCl,total} =M_{HCl} *V_{HCl} =0.111*0.025=2.78x10^{-3} moles$

From the second titration, the moles of excess of HCl is:

$n_{HCl,excess} =n_{NaOH} =M_{NaOH} *V_{NaOH} =0.132*0.01105=1.46x10^{-3} moles$

The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:

$n_{HCl,reacts} =n_{HCl,total}-n_{HCl,excess} =2.78x10^{-3} moles-1.46x10^{-3} moles=1.32x10^{-3} moles$

The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:

$m_{Al(OH)3} =1.32x10^{-3} molesHCl*\frac{1molAl(OH)3}{3molesHCl} *\frac{78gAl(OH)3}{1molAl(OH)3} =0.03g$

The percentage of Al(OH)₃ is:

$Percentage-Al(OH)3=\frac{m_{Al(OH)3} }{m_{antiacid} } *100=\frac{0.03}{0.196} =15.3$ %

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3 years ago