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kogti [31]
3 years ago
13

Negative radical in the periodic table​

Chemistry
1 answer:
AVprozaik [17]3 years ago
7 0

Answer:

Negative Radicals ---→

1).Flouride =  F⁻

2). Chloride = Cl⁻

3) Bromide = Br⁻

4) Iodide = I⁻

5) Sulphate = SO₄²⁻

6) Oxide = O²⁻

7) Nitride = N³⁻

8) Sulphur = S²⁻

9) Carbide = C⁴⁻

10) Hydroxide = OH⁻

11) Nitrate = NO₃⁻

12) Carbonate = CO²⁻

13) Hydrogen Carbonate = HCO₃⁻

14) Sulphate = SO₄²⁻

15).Sulphite = SO₃³⁻

16) Nitrate = NO₃⁻

17)Nitrite = NO₂⁻

18) Hydrogen Sulphite = HSO₃⁻

19) Hydrogen Sulphate = HSO₄⁻

20) Chromate = CrO₄²⁻

There are lot of Radicals which cannot be easily written. Although some of the Important Radicals which are commonly used are mentioned above.

Hope it helps,

Please mark me as the brainliest

Thank you

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See below!

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Aluminum has a +3 charge, and fluorine has a -1 charge.  Since the charge has to be zero, you need three fluorines, giving you AlF₃.

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6 0
3 years ago
Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p
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Answer : The equilibrium constant K_c for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of Br_2.

\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}

\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M

Now we have to calculate the dissociated concentration of Br_2.

The balanced equilibrium reaction is,

                              Br_2(g)\rightleftharpoons 2Br(aq)

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of Br_2 = \alpha = 1.2 %

So, the dissociate concentration of Br_2 = C\alpha=1.731M\times \frac{1.2}{100}=0.2077M

The value of x = 0.2077 M

Now we have to calculate the concentration of Br_2\text{ and }Br at equilibrium.

Concentration of Br_2 = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of Br = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

K_c=\frac{[Br]^2}{[Br_2]}

Now put all the values in this expression, we get :

K_c=\frac{(0.4154)^2}{1.5233}=0.1133

Therefore, the equilibrium constant K_c for the reaction is, 0.1133

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