CHEMISTRY COLLEGE

How many unpaid electrons are in 1s2 2s2 2p6 3s2 3p6

Answers

Answer 1

Answer:

Answer:

As we can see, the first shell has its 2 electrons

the second shell has its 8 electrons

we know that the third shell also needs 8 electrons and we can see that the third shell also has 8 electrons.

Therefore, all the shells of this atom are filled and this atom is stable

So we can say that there are no unpaired electrons

Kindly mark Brainliest, thanks

Answer 2

Answer:

Answer:

No unpaired electrons

Explanation:

This is because it is diamagnetic. Hope that helps:)

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Answers

Answer:

How do u post a picture as the question

A sample of gas has an initial volume of 3.00 L and an initial pressure of 2.14 atm. If the volume expands to 8.15 L, what is the final pressure?a. 0.855 atm b. 0.788 atm c. 3.49 atm d. 5.81 atm

Answers

Answer:

The final pressure is 0.788 atm (option b).

Explanation:

Boyle's law says that the volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure. That is: if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases. This is expressed mathematically as the product of pressure times volume equal to a constant value:

P*V=k

Assuming a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment, by varying the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:

P1*V1=P2*V2

In this case:

P1= 2.14 atm
V1= 3 L
P2= ?
V2= 8.15 L

Replacing:

2.14 atm*3 L= P2* 8.15 L

Solving:

$(2.14 atm*3 L)/(8.15 L) =P2$

0.788 atm= P2

The final pressure is 0.788 atm (option b).

A chemist has two solutions of H2SO4. One has a 40% concentration and the other has a 25% concentration.How many liters of each solution must be mixed to obtain 78 liters of a 28% solution?

liters of the 40% solution and

liters of the 25% solution must be mixed to obtain a 28% solution of H2SO4.

(Round to the nearest tenth, if necessary.)

Answers

Answer:

16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

Explanation:

Let the required volume of solution 1 be represented by x.

The required volume of solution 2 would then be 78-x.

The number of moles of solution 1 that would be required = 0.4x

The number of moles of solution 2 that would be required = 0.25(78-x)

The number of moles of the final mixture = 78 x 0.28 = 21.84

moles of solution 1 + moles of solution 2 = moles of final mixture

0.4x + 0.25(78 - x) = 21.84

0.4x + 19.5 - 0.25x = 21.84

0.4x - 0.25x = 21.84 - 19.5

0.15x = 2.34

x = 15.6 liters

To the nearest tenth = 16 liters

Liters of 40% solution needed = 16 liters

Liters of 25% solution needed = 78 - 16 = 62 liters.

Hence, 16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

Classify each of the following chemical reactions as a synthesis decomposition and single displacement or double displacement reaction

Answers

1. Synthesis reaction : there is only 1 product formed from 2 or more reactant

E.g:

$H_2(g)+N_2(g)\text{ }\Rightarrow2NH_3(g)\text{ }$

2. Decomposition : reaction that occurs in presence of UV light and only 1 reactant that decomposes into 2 or more products.

E.g:

$CH_3Br(g)+UV_(light)\Rightarrow CH_3(g)\text{ + Br (g)}$

3. Single displacement :reaction that occurs when 1 reactant displaces other reactant from its compound:

E.g:

$Zn(s)+CoCl_2(aq)\text{ }\Rightarrow ZnCl_2(aq)\text{ + Co(s)}$

4. Double displacement :reaction that occurs when both reactant displaces each other.

E.g :

$K_2S(aq)+Co(NO_3)_2\Rightarrow2KNO_3(aq)\text{ + }CoS(s)\text{ }$

C6H6NCl On the basis of the molecular formula, 1H NMR data, and IR data provided, propose a consistent structure.H`NMR data: δ 300 (s, 2H), 6.57 (d, 2H), 7.05 (d, 2H)
IR data: 3400 (broad), 3250 (broad), 1590, 820 cm^(-1).

Draw your proposed structure below.

Answers

Answer:

Please refer to the attachment below.

Explanation:

Please refer to the attachment below for explanation.

10. Predict the mass of nitrogen dioxide produced if 2.30 L of ammonia are allowed to reactwith excess oxygen gas at STP?

Answers

Answer:

Mass of nitrogen dioxide produced = 4.6 g

Explanation:

Given data:

Volume of ammonia = 2.30 L

Mass of nitrogen dioxide produced = ?

Solution:

Chemical equation:

4NH₃ + 7O₂ → 4NO₂ + 6H₂O

Number of moles of ammonia at STP:

PV = nRT

n = PV/RT

n = 1 atm × 2.30 L / 0.0821 atm.L/K.mol × 273 K

n = 2.30 atm .L / 22.414 atm.L/mol

n = 0.1 mol

Now we will compare the moles of ammonia with nitrogen dioxide from balance chemical equation.

NH₃ : NO₂

4 : 4

0.1 : 0.1

Mass of NO₂:

Mass = number of moles × molar mass

Mass = 0.1 mol × 46 g/mol

Mass = 4.6 g

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