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Savatey [412]
3 years ago
10

A. Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to t

he element with the most negative electron affinity, Eea?
1. 2s2
2. 2s2 2p2
3. 2s2 2p5
4. 2s2 2p6
B. Arrange the following elements from greatest to least tendency to accept an electron.
Rank from greatest to least tendency to accept an electron. To rank items as equivalent, overlap them.
1. Sr
2. Sn
3. Rb
4. Te
5. I
Chemistry
1 answer:
NeTakaya3 years ago
6 0

Answer:

2s2 2p5

Rb < Sr< Sn< Te<I

Explanation:

Electron affinity is the ability of an atom to accept electrons to form negative ions.

Electron affinity is a periodic trend that decreases down the group but increases across the period.

This accounts for the trends observed in the answer. The atom having the electronic configuration, 2s2 2p5 must be a halogen and it exhibits the highest value of electron affinity.

Also, since electron affinity increases across the period, the electron affinities of the elements increases. Therefore, the arrangement of atoms as shown in the answer depends on increasing electron affinity.

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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
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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%

3 0
3 years ago
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Alenkasestr [34]

Answer:

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