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sweet-ann [11.9K]
2 years ago
14

How do the transition metals in Period 4 affect the pattern of ionization energies in Group 3A(13)? How does this pattern compar

e with that in Group 3B(3)?
Chemistry
1 answer:
serious [3.7K]2 years ago
6 0

The transition metals is Group 3B(13), there is a smoother decrease in ionization energy because these group contains only the transition metals.

The quantity of strength needed to remove an electron from a selected gaseous atom or ion is called the ionisation electricity Group 3B(13) . not just the atoms that are gases at ambient temperature are protected via it; all the elements at the periodic desk are included.

Ionization energy:

the desired energy to put off an outermost electron from a neutral atom is known as ionization electricity.

In typically, up to down inside the periodic table ionization power is decreases but in group 3A and Group 3B(13)  (thirteen) it's miles irregular because the arrival of the transition metals in four length due to this Ga, In and Tl elements indicates better the ionization energies so there's no pattern on this organization.

To learn more about Group 3B(13) refer the link:

brainly.com/question/5489194

#SPJ4

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A hot toluene stream, which has mass flow rate of 8.0 kg/min, is cooled by cooling water in a cocurrent heat exchanger; its temp
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If the spin of one electron in an orbital is clockwise, what is the spin of the other electron in that orbital?
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3 0
3 years ago
The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate
Karolina [17]

<u>Answer:</u> The value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 3.45

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-1=1

Putting values in above equation, we get:

3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084

The equation used to calculate concentration of a solution is:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}

Initial moles of (CH_3)_3CCl(g) = 1.00 mol

Volume of the flask = 5.00 L

So, \text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M

For the given chemical reaction:

                (CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)

Initial:               0.2                    -                        -

At Eqllm:          0.2 - x               x                       x

The expression of K_c for above reaction follows:

K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}

Putting values in above equation, we get:

0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

[(CH_3)_2C=CH]=x=0.094M

[HCl]=x=0.094M

[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M

Hence, the value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

8 0
3 years ago
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