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Tju [1.3M]
2 years ago
11

Which group will have an electron configuration that ends in s2?

Chemistry
1 answer:
Step2247 [10]2 years ago
8 0
<h2>Answer:   c . Alkaline earth metals (Group 2)</h2>

<h3>Explanation:</h3>

Group 2 metals have 2 electrons in their outer shell. These two electrons are usually found in the s orbital, hence the s².

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Which particle diagram represents a mixture of three substances
Anni [7]

Following the key in the diagram (see the attached image), the only particle diagram that represents a mixture of three substances is diagram 2.

To simplify it, let us replace the key in the diagram as follows;

  1. atom of one element = A
  2. atom of different element = B

Diagram 1 consists of only AA and AB

Diagram 2 consists of AA, BB, and AB.

Diagram 3 consists of AA and ABA

Diagram 4 consists of AA and BAB

Thus, only diagram 2 has a mixture of 3 substances.

More on mixtures can be found here: brainly.com/question/6594631

4 0
3 years ago
A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide
Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe

Next, the total moles:

n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol

After that, since the process is isobaric, we can compute the work as:

W=P(V_2-V_1)

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3

Thereby, the magnitude and direction of work turn out:

W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0
3 years ago
. In which reaction is nitric acid acting as an oxidising agent?
Talja [164]

Answer:

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

Explanation:

Hello,

In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.

Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.

Regards.

8 0
3 years ago
Read 2 more answers
What is the main purpose of using NaOH in the saponification reaction . NaOH works as the catalyst b. NaOH solution is the solve
Alex777 [14]

Answer:

C. NaOH acts as a reactant in the reaction

Explanation:

Because during the saponification process, Na+ replaces the H+ in the fatty acid been used for the saponification process

7 0
3 years ago
g The reaction; 4 Ag(s) + O2(g) ----&gt; 2 Ag2O(s), is exothermic. Which statement about the reaction is correct? (A) It is spon
melomori [17]

Answer:

The correct answer is B. It is spontaneous only at low temperatures.

Explanation:

In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.

The spontaneity of a reaction is given by the equation:

ΔG = ΔH - TΔS

where:

ΔH: enthalpy variation

T: absolute temperature

ΔS: entropy variation

As the reaction is exothermic, ΔH<0

As the reaction order increases (the reagents are solid and gas and their product is solid), ΔS<0

Therefore, the reaction will be spontaneous when ΔG is negative.

ΔG = ΔH - TΔS

That is, the entropy term must be smaller than the enthalpy term.

Hence, the reaction will be spontaneous only at low temperatures.

4 0
3 years ago
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