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saveliy_v [14]
3 years ago
6

Consider the elements: Na, Mg, Al, Si, P.

Chemistry
1 answer:
Olegator [25]3 years ago
8 0

Answer:

The elements mentioned in the series are Na, Mg, Al, Si, and P. It can be seen that all these elements are located in the same period. The atomic number of the mentioned elements are Na-11, Mg-12, Al-13, Si-14, and P-15.

a) There will be an increase in the ionization energy with the increase in the element's atomic number across a period. More energy is needed to withdraw an electron from a completely occupied shell in comparison to an incompletely occupied shell.

The atomic number of Na is 11. When one electron is withdrawn from Na, it gets converted into the inert gas configuration of Ne. Thus, it will require more energy to withdraw the second electron from Na. Hence, Na exhibits the lowest second ionization energy.

b) Across the period, with an increase in the element's atomic number, the atomic radii reduces from left to right. Thus, P exhibits the smallest atomic radius.

c) The metallic nature of the elements reduces from left to right across a period in the periodic table. Thus, P is the least metallic element.

d) Diamagnetic signifies towards the element that exhibits pair electrons in its sub-shells. The electronic configuration of Mg is,

1s2 2s2 2p6 3s2

In Mg, no unpaired electrons are present, while all the remaining elements mentioned exhibit unpaired electrons in their valence shell. Thus, Mg is the diamagnetic element.

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Given the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Lelechka [254]

Answer:

ΔG°rxn = -69.0 kJ

Explanation:

Let's consider the following thermochemical equation.

N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ

Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.

3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ

8 0
3 years ago
A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

  • Partial pressure of methane (pCH₄): 431 mmHg
  • Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 0
2 years ago
KHP is a monoprotic acid which provides one H+ ion. How would your results be affected if a diprotic acid (such as sulfuric acid
kari74 [83]

Explanation:

A.

In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.

B.

Equation of the reaction

2NaOH + H2SO4 --> Na2SO4 + 2H2O

Number of moles of H2SO4 = molar concentration × volume

= 0.75 × 0.0105

= 0.007875 moles.

By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875

= 0.01575 moles.

Molar concentration of NaOH = number of moles ÷ volume

= 0.01575 ÷ 0.0175

= 0.9 M of NaOH.

7 0
3 years ago
How many neon atoms are contained in (7.27x10^2) grams of neon?
statuscvo [17]

Answer: 2.19x10^25 atoms

Explanation:

Molar Mass of Neon = 20g/mol

1mole(20g) of Neon contains 6.02x10^23 atoms.

Therefore, 7.27x10^2g of Neon will contain x atoms i.e

X atoms = (7.27x10^2x6.02x10^23)/20 = 2.19x10^25 atoms

4 0
3 years ago
What is the molarity of a stock solution if 60 mL were used to make 150 ml<br>of a .5M solution? ​
olga2289 [7]

The molarity of the stock solution is 1.25 M.

<u>Explanation:</u>

We have to find the molarity of the stock solution using the law of volumetric analysis as,

V1M1 = V2M2

V1 = 150 ml

M1 = 0.5 M

V2 = 60 ml

M2 = ?

The above equation can be rearranged to get M2 as,

M2 = $\frac{V1M1}{V2}

Plugin the values as,

M2 = $\frac{150 \times 0.5}{60}

       = 1.25 M

So the molarity of the stock solution is 1.25 M.

4 0
3 years ago
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