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Shtirlitz [24]
3 years ago
6

For which of the following elements would the most common ion be expected to have a larger radius than that of its corresponding

atom?
1. Cl
2. Na
3. Cu
4. Sr
Chemistry
1 answer:
Andre45 [30]3 years ago
8 0

<u>Answer:</u> The ion that is expected to have a larger radius than the corresponding atom is chlorine.

<u>Explanation:</u>

There are two types of ions:

  • <u>Cations:</u> They are formed when an atom looses its valence electrons. They are positive ions.
  • <u>Anions:</u> They are formed when an atom gain electrons in its outermost shell. They are negative ions.

For positive ions, the removal of electron increases the nuclear charge for an outermost electron because the outermost electrons are more strongly attracted by the nucleus. So, the effective nuclear charge increases for cations and thus, the size of the cation will be smaller than that of the corresponding atom.

For negative ions, the addition of electron decreases the nuclear charge for an outermost electron because the outermost electrons are less strongly attracted by the nucleus. So, the effective nuclear charge decreases for anions and thus, the size of the anion will be larger than that of the corresponding atom.

For the given options:

<u>Option a:</u> Chlorine

Chlorine gains 1 electron and form Cl^- ion

<u>Option b:</u> Sodium

Sodium looses 1 electron and form Na^+ ion

<u>Option c:</u> Copper

Copper looses 2 electrons and form Cu^{2+} ion

<u>Option d:</u> Strontium

Strontium looses 2 electrons and form Sr^{2+} ion

Hence, the ion that is expected to have a larger radius than the corresponding atom is chlorine.

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</span>
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Dima020 [189]

Answer:

Hey There!

Hope it helps

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Read 2 more answers
In an experiment, a 0.4351 g sample of benzil (C14H10O2) is burned completely in a bomb calorimeter. The calorimeter is surround
andre [41]

Answer:

The enthalpy change during the reaction is -7020.09 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 994.1 J/^oC

T_{final} = final temperature = 27.60^oC

T_{initial} = initial temperature = 25.10^oC

Now put all the given values in the above formula, we get:

q=994.1 J/^oC\times (27.60-25.10)^oC

q= 2,485.25 J

The heat gained by water present in calorimeter. = q'

q'=mc\times (T_{final}-T_{initial})

where,

q' = heat gained = ?

m = mass of water = 1.153\times 10^3 g

c' = specific heat of water = 4.184 J/^oC

T_{final} = final temperature = 27.60^oC

T_{initial} = initial temperature = 25.10^oC

q'=1.153\times 10^3 g\times 4.184 J/^oC\times (27.60-25.10)^oC

q ' = 12,060.38 J

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{Q}{n}

where,

\Delta H = enthalpy change = ?

Q = heat gained = -(q+q') = -(2,485.25 J + 12,060.38 J)= -14,545.63 J

Q = -14.54563 kJ

n = number of moles fructose = \frac{\text{Mass of benzil}}{\text{Molar mass of benzil}}=\frac{0.4351 g}{210 g/mol}=0.002072 mole

\Delta H=-\frac{-14.54563 kJ}{0.002072 mole}=-7020.09 kJ/mole

Therefore, the enthalpy change during the reaction is -7020.09 kJ/mole.

3 0
3 years ago
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