3 years ago

Is my answer right?

Chemistry

1 answer:

Ahat [919]3 years ago

7 0

Yes the answer is 1s22s22p63s23p64s23d5

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Ascorbic acid (vitamin

svlad2 [7]

Let us see the structure of ascorbic acid

As shown there is no COOH group however the OH group can lose a proton and forms conjugate base

The conjugate base formed is stabilized due to resonance

More the stability of conjugate base more the strength of acid

Hence ascorbic acid behaves as an acid

8 0

3 years ago

4 A Small rock

vodomira [7]

3 millimeters is the rock

6 0

3 years ago

During the formation of an ionic compound, the element with the higher ionization energy will gain electrons.

Lesechka [4]

False because you have to take it out and do it right

8 0

3 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

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

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

2 years ago

A gas has a volume of 5.0 L at a pressure of 50 kPa. What happens to the volume when temperature is held constant and the pressu

Leto [7]

2.0L is the answer for sure

3 0

3 years ago