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3 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]3 years ago

5 0

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

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B. Determine the percent composition of each element in NaCl. (1 point) Must show work to

andrey2020 [161]

Answer:

Sodium (Na): (.5 point)

22.99+35.453=58.433. 22.99/58.433= 39%

Chlorine (Cl): (.5 point)

22.99+35.453=58.433. 35.453/58.433= 61%

Explanation:

8 0

2 years ago

Cuántos electrones de Valencia tiene Ga

Zinaida [17]

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1

Primero realizas la configuración electrónica que es la que te puse allá arriba.

Después miras el nivel en que termina, puede ser 1, 2, 3, 4 etc.

Entonces como el último número de la configuración electrónica es 4, entonces ese es el nivel

Y los electrones de el último nivel son los de Valencia

4s2, 4p1 sumas 2+1 que son los electrones que se encuentran en el último nivel.

por eso hay 3 electrones de valencia.

4 0

3 years ago

Find the volume of hydrogen gas formed when 1.5g of aluminum reacts with aq NaOH at 27 degrees Celcius.

Rudiy27

Answer:

how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?

(A) n=m/M,

n(Al)=5.4/27=0.2 moles

n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles

Number of oxygen atoms= n(O2)*Avogadro's number

=0.15*6.02*10^23=9.03*10^22 oxgyen atoms

(B)

n=m/M

n(Al)=0.6/27=0.02222 moles

n(O2)=n(Al)*3/4=0.016666 moles

m=n*M

m(O2)=0.0166666*32=0.53333 grams

8 0

2 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago

Why a mixture of magnesium hydroxide and water is known as milk of magnesia

gregori [183]

because of its milk-like appearance. ... Since the dissociation of this small amount of dissolved magnesium hydroxide is complete, magnesium hydroxide is considered a strong electrolyte. Its low solubility makes it a weak base.

3 0

3 years ago