All categories

3 years ago

Plz solve mcqs#04 with full detailed.note

Chemistry

1 answer:

Sholpan [36]3 years ago

8 0

Answer:

The correct option is;

(B) 1 s², 2s², 2p⁶, 3s², 3p³

Explanation:

The electron configuration is the outline of the electron arrangement about a nucleus

In the systemic pattern of electron arrangement within an atom, there are, s, p, d, f orbitals

The maximum number of electrons in an s, p and d orbital = 2, 6, and 10 respectively

Based on Aufbau's principle the electrons are arranged based on the order of their energy level

The charge is presented by the number of electrons in the outermost shell, an element able to form an ion of charge of -3 will gain 3 electrons to complete its outermost shell

Among the options given, option B is the only option that has the capacity to take the electrons to complete the number of electrons in the p orbital outermost shell to 6 from 3, that is 3p³ + 3e⁻→ 3p⁶.

You might be interested in

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 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)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

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:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

What is the concentration of 60 mL of H3PO4 if it is neutralized by 225 mL of 2 M Ba(OH)2?

goldfiish [28.3K]

7.5 M is the concentration of 60 ml of H3PO4 if it is neutralized by 225 ml of 2 M Ba(OH)2.

Explanation:

Data given:

volume of phosphoric acid, Vacid =60 ml

volume of barium hydroxide, Vbase = 225 ml

molarity of barium hydroxide, Mbase = 2M

Molarity of phosphoric acid, Macid =?

the formula for titration is used as:

Macid x Vacid = Mbase x Vbase

rearranging the equation to get Macid

Macid = $\frac{Mbase x Vbase}{Vacid}$

Macid = $\frac{225 X 2}{60}$

Macid = 7.5 M

the concentration of the phosphoric acid is 7.5 M and the volume is 60 ml. Thus 7.5 M solution of phosphoric acid is used to neutralize the barium hydroxide solution of 2M.

8 0

4 years ago

Read 2 more answers

Compute the freezing point of this solution:

likoan [24]

Answer:

Freezing point = 1.25

Explanation:

If we increase the concentration of the solution, the concentration of H+ does not change.

Convert 2.5% in to decimal

2.5% = 2.5 ÷100

= 0.025

The freezing point = 0.025 × 50

= 1.25

8 0

3 years ago

what volume of a 0.138 m potassium hydroxide solution is required to neutralize 26.0 ml of a 0.205 m nitric acid solution?

Law Incorporation [45]

A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is ( $M1V1 = M2V2$ ).

Utilize the titration method of $M1V1 = M2V2$ in view that we're given the concentrations of every compound and the quantity of $KOH$ . Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.

M1 = initial mass
V1= initial volume
M2 = final mass
V2= final volume

$(M1V1 = M2V2)$
(0.138)(V1) = (0.205)x(26.0)
V2=(0.205)x(26.0)\ 0.138
V2 = 47.10 M/L
The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L