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3 years ago

14

In which orbitals would the valence electrons for selenium (Se) be placed?

2 answers:

RSB [31]3 years ago

8 0

Selenium falls in the same column with oxygen therefore it has the same number of valence electrons with oxygen which is 6. It fills the 3d orbital but only 4s and 4p are counted. The electronic configuration is [Ar]3d^10 4s^2.

elixir [45]3 years ago

8 0

Selenium falls in the same column with oxygen therefore it has the same number of valence electrons with oxygen which is 6. It fills the 3d orbital but only 4s and 4p are counted. The electronic configuration is [Ar]3d^10 4s^2

Explanation:

Selenium has 6 valence electrons, two of which are placed in the S orbital and 4 of which are placed in the P orbital. In its electron structure, it is quite comparable to oxygen.

At the first energy level, the only orbital accessible to electrons is the 1s orbital, but at the next level, as well as a 2s orbital, there are 2p orbitals. A p orbital is formed like 2 identical balloons tied commonly at the nucleus. The orbital shows where there is a 95% chance of getting a critical electron.

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A sample of oxygen occupies 20.1 liters under a pressure of 1520 torr at 25.0o What volume would it occupy at 25.0oC if the pres

Zolol [24]

Answer:

The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L

Explanation:

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant, so that the pressure of a gas in a closed container is inversely proportional to the volume of the container. That is, if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Considering an initial state 1 and a final state 2, it is true:

P1* V1= P2*V2

In this case:

P1= 20.1 L
V1= 1520 torr
P2= 760 torr
V2= ?

Replacing:

20.1 L* 1520 torr= 760 torr* V2

Solving:

$V2=\frac{20.1 L* 1520 torr}{760 torr}$

V2= 40.2 L

<em><u>The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L</u></em>

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4 0

2 years ago

What do I do if I like someone but they think your annoying

Mrrafil [7]

Answer:

Say sorry, and then send them a gift.

Explanation:

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2 years ago

Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in

IgorC [24]

Answer:

669.48 kJ

Explanation:

According to the question, we are required to determine the heat change involved.

We know that, heat change is given by the formula;

Heat change = Mass × change in temperature × Specific heat

In this case;

Change in temperature = Final temp - initial temp

= 99.7°C - 20°C

= 79.7° C

Mass of water is 2000 g ( 2000 mL × 1 g/mL)

Specific heat of water is 4.2 J/g°C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

= 669,480 joules

But, 1 kJ = 1000 J

Therefore, heat change is 669.48 kJ

7 0

3 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

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3 years ago

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nika2105 [10]

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3 years ago

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