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3 years ago

What is the unabbreviated electron configuration for Bromine?

Chemistry

2 answers:

Komok [63]3 years ago

8 0

Answer:

[Ar] 3d¹⁰ 4s² 4p⁵

Explanation:

hope it helps

xz_007 [3.2K]3 years ago

4 0

Answer:

[Ar] 3d¹⁰ 4s² 4p⁵

Explanation:

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How would decreasing the volume of the reaction vessel affect each of the following equilibria? 2NOBr(g) --> 2NO(g)+Br2(g)

Alex_Xolod [135]

As per Le Chatelier principle, when a system in equilibrium is disturbed, the reaction will try to compensate the change to restore the equilibrium.

This reaction occurs in gas phase, so the volume is inversely proportional to the pressure.

Then a decrease in volume will cause an increase in pressure, so the system will tend to react in the direction that compensates this increase, this is the system will try to reduce the number of moles of particles to reduce the increase of the pressure.

As you see, there are 3 particles of products (2 of NO and 1 of Br) for every 2 particles of reactant (NOBr).

That means, that the equilibrium will displace to the left, this is the concentration of NOBr will increase while the concentration of NO and Br will decrease.

3 0

3 years ago

The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq

Pavlova-9 [17]

Answer : The molarity of $I_3^-$ in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

$2S_2O_2^{-3}(aq)+I_3(aq)\rightarrow S_4O_2^{-6}(aq)+3I^-(aq)$

First we have to calculate the moles of $Na_2S_2O_3$ .

$\text{Moles of }Na_2S_2O_3=\text{Molarity of }Na_2S_2O_3\times \text{Volume of solution}$

$\text{Moles of }Na_2S_2O_3=0.260mole/L\times 0.0296L=0.007696mole$

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of $I_3^-$ .

From the balanced chemical reaction, we conclude that

As, 2 moles of $S_2O_2^{-3}$ react with 1 mole of $I_3^-$

So, 0.007696 moles of $S_2O_2^{-3}$ react with $\frac{0.007696}{2}=0.003848$ mole of $I_3^-$

The moles of $I_3^-$ = 0.003848 mole

Now we have to calculate the molarity of $I_3^-$ .

$\text{Molarity of }I_3^-=\frac{\text{Moles of }I_3^-}{\text{Volume of solution}}$

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

$\text{Molarity of }I_3^-=\frac{0.003848mole}{0.03L}=0.128mole/L=0.128M$

Therefore, the molarity of $I_3^-$ in the solution is, 0.128 M

8 0

3 years ago

What kind of intermolecular forces act between a hydrogen cyanide molecule and a chloromethane molecule? note: if there is more

solmaris [256]

London dispersion forces, dipole forces

3 0

3 years ago

When a metal was exposed to light at a frequency of 4.07× 1015 s–1, electrons were emitted with a kinetic energy of 3.30× 10–19

Licemer1 [7]

Answer : The maximum number of electrons released = $1.432\times 10^{12}electrons$

Explanation : Given,

Frequency = $4.07\times 10^{15}s^{-1}$

Kinetic energy = $3.30\times 10^{-19}J$

Total energy = $3.39\times 10^{-7}J$

First we have to calculate the work function of the metal.

Formula used :

$K.E=h\nu -w$

where,

K.E = kinetic energy

h = Planck's constant = $6.626\times 10^{-34}J/s$

$\nu$ = frequency

w = work function

Now put all the given values in this formula, we get the work function of the metal.

$3.30\times 10^{-19}J=(6.626\times 10^{-34}J/s\times 4.07\times 10^{15}s^{-1})-w$

By rearranging the terms, we get

$w=2.367\times 10^{-18}J$

Therefore, the works function of the metal is, $2.367\times 10^{-18}J$

Now we have to calculate the maximum number of electrons released.

The maximum number of electrons released = $\frac{\text{ Total energy}}{\text{ work function}}$

The maximum number of electrons released = $\frac{3.39\times 10^{-7}J}{2.367\times 10^{-19}J}=1.432\times 10^{12}electrons$

Therefore, the maximum number of electrons released is $1.432\times 10^{12}electrons$

8 0

3 years ago

Consider four elements from Group 17: fluorine in the second period, chlorine in the third period, bromine in the fourth period,

Ghella [55]

<span>Fluorine
For the group 17 elements (Halogens), the ionization energy decreases as the period increases (the elements become less reactive). And since of the 4 elements given, fluorine is from the lowest period, it has the highest ionization energy.</span>

7 0

3 years ago

Read 2 more answers