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Luden [163]
3 years ago
5

The first four ionization energies for an imaginary element, Xz, are E1 = 102 kcal, E2 = 186 kcal, E3 = 4021 kcal, and E4 = 4862

kcal. The number of valence electrons most probable as indicated by this ionization data is:
Chemistry
2 answers:
baherus [9]3 years ago
7 0

Answer: Two valence electrons.

Explanation: Ionization energy is the energy required to remove electrons from valence shell of an atom in its gaseous state.

More closer are the electrons to the nucleus, more energy is required to remove them. Ionization energy increases as the effective nuclear charge increases.

First two ionization energy values are low and there isn't a much difference between them but the third ionization energy is very much high as compared to first and second ionization energy values. The fourth ionization energy value is also closer to third ionization energy value.

It indicates that first and second electrons are removed from same shell. Similarly, third and fourth electrons are also removed from same shell which is different than the shell from which first and second electrons are removed.

So, outer most shell that is valence shell has only two electrons. Third ionization energy has increased a lot as the third electron is removed from the inner shell and first two electrons are removed from valence shell.

Anni [7]3 years ago
4 0
Hi :) If your answers choices are what I'm thinking, the answer is two :)
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Why do most atoms form chemical bonds?
zloy xaker [14]

Answer:

Why do most atoms form chemical bonds? They want a full outer shell of electrons, so the lose, gain, or share electrons with other elements, forming compounds, until they have 8 valence electrons and become stable. Double and triple covalent bonds that have greater bond energy and are shorter than single bonds.

Explanation: HOPE THIS HELPS YOU..

3 0
3 years ago
At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank
hodyreva [135]

Answer:

\large \boxed{\text{-92 $^{\circ}$C}}

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}

2. Temperature in degrees Celsius

\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}

8 0
3 years ago
Lewis structure for 2O2
Mrrafil [7]

the big number describes the number ratio in a chemical equation

so for example,

2H2 + O2 --> 2H2O means

2 moles of hydrogen reacts with one mole of oxygen to form 2 moles of water

and as you know, the small (subscript) number determines the number of atoms of that element in one molecule of a compound

so I believe that drawing a normal lewis structure ( O=O ) should be correct

6 0
3 years ago
Read 2 more answers
Convert 1 bromopropane to bromoethane.​
Dovator [93]

Answer:

The Dehydrohaogenation of 1-bromo propane with alcoholic KOH gives propene which on again hydrohalogenation with HBr gives 2-bromo propane due to Markonikove's rule for addition.

Explanation:

6 0
3 years ago
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.
yaroslaw [1]

Answer:

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

Explanation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

initially

3.0 atm      0              0

At equilibrium

(3.0-2p)     p               3p

Equilibrium partial pressure of nitrogen gas = p = 0.90 atm

The expression of a pressure equilibrium constant will be given by :

K_p=\frac{p_{N_2}\times (p_{H_2})^3}{(p_{NH_3})^2}

K_p=\frac{p\times (3p)^3}{(3.0-2p)^2}

=\frac{0.90 atm\times (3\times 0.90 atm)^3}{(3.0-2\times 0.90 atm)^2}

K_p=12.30\approx 12.

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

3 0
3 years ago
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