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Marta_Voda [28]
3 years ago
8

as the elements period 3 are considered in order of increasing atomic number, the number of principal energy levels in each succ

essive element
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

stay the same.

Explanation:  Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):

Na:  1s2 2s2 2p6 3s1

Ar:    s2 2s2 2p6 3s2 3p6

As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:

Na: 3s1

Ar: 3s2, 3p6

All the new electrons electrons filled the third energy level (3s and 3p).  So the energy level does not change, just the orbitals.

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Each carbon atom is covalently bonded to 4 other carbon atoms in diamond. A large amount of energy is required to split these atoms apart. This is because of the fact that covalent bonds are strong.

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Compare and contrast the general properties of alloys and pure metals
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Ethanol has a heat of vaporization of 38.56kj/mol and a normal boiling point of 78.4 ∘c.
Elanso [62]

This is an incomplete question, here is a complete question.

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 14 °C?

Answer : The vapor pressure of ethanol at 14.0^oC is 5.174\times 10^{-2}atm

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 14.0^oC = ?

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm

T_1 = temperature of ethanol = 14.0^oC=273+14.0=287K

T_2 = normal boiling point of ethanol = 78.4^oC=273+78.4=351.4K

\Delta H_{vap} = heat of vaporization = 38.56 kJ/mole = 38560 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{38560J/mole}{8.314J/K.mole}\times (\frac{1}{287K}-\frac{1}{351.4K})

P_1=5.174\times 10^{-2}atm

Hence, the vapor pressure of ethanol at 14.0^oC is 5.174\times 10^{-2}atm

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