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bazaltina [42]
3 years ago
15

Atoms with loosely held valence electrons have 1. high ionization energy and high electron affinity. 2. high ionization energy a

nd low electron affinity. 3. low ionization energy and high electron affinity. 4. low ionization energy and low electron affinity. 017
Chemistry
1 answer:
EastWind [94]3 years ago
3 0

Answer:

4

Explanation:

Ionization energy can be defined as the energy required for an atom to lose its valence electron to form an ion. Hence, it deals with how easily an atom would lose its electron and form an ion. As the valence electrons are lossless bound to the outermost shell, they can easily be lost without much problem or better still they can be lost easily. Hence, the energy change here is small and thus we can conclude that the ionization energy here is low.

The electron affinity works quite differently from the ionization energy. It deals with the way in which a neutral atom attracts an electron to form an ion. For an electron with loose valence electrons, the sure fact is that it does not really need these electrons. Hence, there is no need for an high electron affinity on its part. Thus, we conclude that the electron affinity is also low

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3 years ago
O que são processos físicos de separação das misturas? Dê 3 exemplos
chubhunter [2.5K]

Answer:

Explanation:

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8 0
3 years ago
When benzene (c6h6) reacts with bromine (br2) bromobenzene(c6h5br) is obtained: c6h6 + br2 → c6h5br + hbr what is the theoretica
saveliy_v [14]

Answer A) : We have to calculate the number of moles of Benzene involved in the reaction,

30 g / 78 moles of benzene = 0.384 moles


For bromine it will be the same process,

65 g / 159.8 moles = 0.406 moles


By observing the reaction given above we can say that the reaction ratio of bromine and benzene is 1 : 1


We need to find the mass of bromobenzene,

which should be, 6(12) + 5 (1)+ 79.90 = 156.9 g/mol


So, the mass of bromobenzene will be 156.9 g/mol X 0.3846 mol = 60.343 g


Hence the theoretical yield will be 60.34 g


Answer B) : To calculate the actual yield we have to divide it with theoretical yield.


(56.7g / 60.343 g ) X100% = 93.96 %


Here, we can say that we got 93.96 % of actual yield.


As we know it is impossible to get 100% yield in any reaction.

3 0
3 years ago
What geometry does VSEPR predict for the central atom in PF5
ddd [48]

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8 0
3 years ago
How many grams are in 1.6 moles of potassium bromide?​
slava [35]

Answer:

190.4g

Explanation:

1.6mol of KBr (119.002g KBr/1 mol) = 190.4g

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6 0
3 years ago
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