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2 years ago

Type in the correct values to correctly represent the valence electron configuration of magnesium: AsB A = B =

2 answers:

7 0

<h2>Answer : The correct answer will be A =3 and B =2 as the electron configuration of Mg will be as $[Ne] 3s^{2}$ where A is 3 and B is 2.</h2><h3>Explanation : </h3>

Magnesium has atomic number as 12. The complete electron configuration will be as $1s^{2}2s^{2}2p^{6}3s^{2}$ .

which can be abbreviated as $[Ne] 3s^{2}$ as the electron configuration resembles to that of neon element.

In the outermost shell the electron in 3 s orbital is only 2. So, therefore 3 is the orbital of S and there are 2 electrons in it.

Alexxandr [17]2 years ago

5 0

The correct answer would be,
A= 3
B=2

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Consider a cell membrane at 298K which has glucose at the concentration of 200mM inside the cell and 20 mM outside the cell, wit

OleMash [197]

Answer:

D. +5.7 kJ/mol

Explanation:

Molar free energy (ΔG) in the transportation of uncharged molecules as glucse through a cell membrane from the exterior to the interior of the cell is defined as:

ΔG = RT ln C in / C out

knowing R is 8,314472 kJ/molK; T is 298K Cin = 200mM and Cout = 20mM

ΔG = 5,7 kJ/mol

Right answer is:

D. +5.7 kJ/mol

I hope it helps!

4 0

2 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

6 0

2 years ago

At 400 K, the equilibrium constant for the reaction Br2 (g) + Cl2 (g) ↔ 2BrCl (g) is Kp = 7.0. A closed vessel at 400 K is charg

xz_007 [3.2K]

Answer:

(a) The equilibrium partial pressure of BrCl (g) will be greater than 2.00 atm.

Explanation:

Q is the coefficient of the reaction and is calculated the same of the way of the equilibrium constant, but using the concentrations or partial pressures in any moment of the reaction, so, for the reaction given:

Q = (pBrCl)²/(pBr₂*pCl₂)

Q = 2²/(1x1)

Q = 4

As Q < Kp, the reaction didn't reach the equilibrium, and the value must increase. As we can notice by the equation, Q is directly proportional to the partial pressure of BrCl, so it must increase, and be greater than 2.00 atm in the equilibrium.

The partial pressures of Br₂ and Cl₂ must decrease, so they will be smaller than 1.00 atm. And the total pressure must not change because of the stoichiometry of the reaction: there are 2 moles of the gas reactants for 2 moles of the gas products.

Because is a reversible reaction, it will not go to completion, it will reach an equilibrium, and as discussed above, the partial pressures will change.

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