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2 years ago

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Which of the following could be the subshell representation of an element of Group IV?

1 answer:

Afina-wow [57]2 years ago

7 0

The subshell representation of an element of Group IV is option C. s² p², since it contains 4 valence electrons.

<h3>What are Groups?</h3>

In the Periodic Table, the Groups are the columns. For the Main-group elements, the group number coincides with the number of electrons in the valence shell.

Which of the following could be the subshell representation of an element of Group IV?

A. s². No, since there are 2 valence electrons.

B. s² p³. No, since there are 5 valence electrons.

C. s² p². Yes, since there are 4 valence electrons.

D. s² p¹. No, since there are 3 valence electrons.

The subshell representation of an element of Group IV is option C. s² p², since it contains 4 valence electrons.

Learn more about Groups here: brainly.com/question/24144303

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Ray Of Light [21]

I think that it is D because the higher the molecular mass, the higher it's boiling point. I hope that helped:) Bye!

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3 years ago

Which of the following covalent bonds is the most polar?

Bezzdna [24]

Answer:

D.) H-O

Explanation:

Polarity is determined based on the difference in electronegativity of the atoms. The greater the difference, the more polar the bond. The general trend is that the atoms in the top-right corner of the periodic table are the most electronegative.

A.) is incorrect because H-H has no electronegativity difference, making it nonpolar.

B.) and C.) are incorrect because their electronegativity differences are not the greatest.

D.) is correct because the electronegativity difference between the H and O is the greatest.

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2 years ago

15.3 g of nano3 were dissolved in 100g of water in a calorimeter. The temperature of the water drops from 25.00°c to 21.56°c. Ca

Arturiano [62]

Answer:

0.259 kJ/mol ≅ 0.26 kJ/mol.

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 100.0 g).

c is the specific heat of water (c of ice = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).

<em>∵ Q = m.c.ΔT</em>

∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.

<em>∵ ΔH = Q/n</em>

n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.

∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.

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3 years ago

What made alchemy unscientific in its practices?

anzhelika [568]

(B) because it couldn’t turn in to good

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3 years ago

Read 2 more answers

It takes 330 joules of energy to raise the temperature of 24.6 gbenzene from 21 degrees Celsius to 28.7 degrees Celsius at const

lilavasa [31]

Given :

Energy , E = 330 J .

Initial temperature , $T_i=21^oC$ .

Final temperature , $T_f=24.6^oC$ .

Mass of benzene , m = 24.6 g .

To Find :

The molar hear capacity of benzene at constant pressure .

Solution :

Molecular mass of benzene , M = 78 g/mol .

Number of moles of benzene :

$n=\dfrac{24.6}{78} \ mol\\\\n=0.32 \ mol$

Energy required is given by :

$q=nC_p\Delta T\\\\330=0.32\times C_p\times (28.7-21)\\\\C_p=\dfrac{330}{0.32\times 7.7}\ J\ mol^{-1}^oC^{-1} \\\\C_p=133.9\ J\ mol^{-1}^oC^{-1}$

Hence , this is the required solution .

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3 years ago