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bixtya [17]
2 years ago
15

Which of the following notations is the correct noble gas configuration for Li

Chemistry
2 answers:
Anettt [7]2 years ago
8 0

Noble gas configuration for Li : [He]2s¹

<h3>Further explanation </h3>

In an atom, there are levels of energy in the shell and sub-shell

This energy level is expressed in the form of electron configurations.

Lithium with atomic number 3, then the electron configuration:

1s²2s¹

And for noble gas configuration or it can be called Condensed electron configurations :

[He]2s¹

wlad13 [49]2 years ago
8 0

Answer:

[He]2s¹

Explanation:

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Use the ideal gas law:

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so, T = PV / nR

n=0.5
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R = 1/12
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Put the values:

T = PV / nR
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Explain why beryllium is produced when potassium is heated with beryllium
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Explanation:

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Since, potassium is more reactive than beryllium. When potassium reacts with beryllium choride, it replaces beryllium and forms potassium chloride and produces beryllium.

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2 years ago
Which statement about London’s desperation forces is true
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8 0
3 years ago
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
2 years ago
Consider the reaction. 2c(s,graphite) h2= c2h2(g) Based on the equation and the information in the table, what is the enthalpy o
xxTIMURxx [149]
<h3>The enthalpy of the reaction : C.226.73 kJ</h3><h3>Further explanation</h3>

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

Delta H reaction (ΔH) is the difference between product enthalpy and reactant enthalpy (Hp-Hr)

Reaction :

2C + H2 -> C2H2

(ΔH) reaction = (ΔH) product - (ΔH) reactants

(ΔH) reaction = (ΔH) C2H2 - ((ΔH) C + (ΔH) H2)

for free element (ΔH) it has zero value, so (ΔH) C and (ΔH) H2 have zero value

From the available data: Hf (kJ / mol) C2 H2 (g)=  226.73

then

(ΔH) reaction = ΔH) C2H2 - ((ΔH) C + (ΔH) H2)

(ΔH) reaction = 226.73 - (0 + 0)

(ΔH) reaction = 226.73 (answer C)

<h3>Learn more</h3>

(ΔH) reaction

brainly.com/question/1889660

Keywords : (ΔH) reaction, C2H2

8 0
3 years ago
Read 2 more answers
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