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2 years ago

What is the valence number of an element that has 3 electron in its outermost shell?

1 answer:

5 0

Lithium has 3 electrons --- 2 in the first shell, and 1 in the second shell (so one valence electron). Beryllim has 4 electrons --- 2 in the first shell, and 2 in the second shell (so two valence electrons). Boron has 5 electrons --- 2 in the first shell, and 3 in the second shell (so three valence electrons).Jul 27, 2019
Hope that helps and may I get brainliest?

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Which of the following atoms would bond covalently? A. Silicon and sulfur B. Carbon and neon C. Lithium and fluorine D. Calcium

lord [1]

Answer:

For example water is a covalent compound which is formed by the covalent bonding between hydrogen and oxygen atom.

Explanation:

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2 years ago

Describe the size, content,<br> and the density of an atom’s nucleus.

german

Hmm trick question here's my answer: the content: protons charge the electrons into its usage , the size: 1.6fm if I'm wrong sorry if I'm right: HA IN YOUR FACE HA!!!

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2 years ago

How many total orbitals are within the 2s and 2p sublevels of the second

mojhsa [17]

Answer:4

Explanation:

Apex

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3 years ago

Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propa

Mademuasel [1]

Answer : The vapor pressure of propane at $25.0^oC$ is 17.73 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 propane at $25.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $25.0^oC=273+25.0=298.0K$

$T_2$ = normal boiling point of propane = $-42.04^oC=230.96K$

$\Delta H_{vap}$ = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

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

$\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})$

$P_1=17.73atm$

Hence, the vapor pressure of propane at $25.0^oC$ is 17.73 atm.

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2 years ago

What type of attraction is the basis for hydrogen bonds (imf)?

Crazy boy [7]

Hydrogen bonds are strong intermolecular forces created when a hydrogen atom bonded to an electronegative atom approaches a nearby electronegative atom. Greater electronegativity of the hydrogen bond acceptor will lead to an increase in hydrogen-bond strength.
I hope this helps

5 0

3 years ago