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2 years ago

How many electrons does fluorine have? How many electron shells? How many electrons are needed to fill the valence shell?

Chemistry

1 answer:

Alex17521 [72]2 years ago

7 0

Answer:

9 electrons when there is no charge shown. It has 2 shells, this is easy to see on the periodic table because it is in the second row. Otherwise you can count. There is 2 electrons on the inner shell then every shell after holds 8. Fluorine only has 9 electrons so 2 for the first shell and 7 on the outter shell. It needs 1 more electron to fill its outter (valence) shell.

Explanation:

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The Molar mass(M.w) in g/mol of 6.3 grams of an ideal gas that placed in 5.0 L tank at ST

BigorU [14]

Answer:

28 g/mol, N2

Explanation:

Given data:

Volume of gas = 5.0 L

Mass of gas = 6.3 g

Pressure = 1 atm

Temperature = 273 K

Molar mass of gas = ?

Solution:

We will calculate the density first.

d = mass/ volume

d = 6.3 g/ 5.0 L

d = 1.26 g/L

Molar mass:

d = PM/RT

M = dRT/P

M = 1.26 g/L× 0.0821 atm.L/mol.K × 273 K/ 1 atm

M = 28 g/mol

Molar mass of N₂ is 28 g/mol thus given gas is N₂.

7 0

2 years ago

Matter is made up of small

Lena [83]

Answer:

atoms and molecules

Explanation:

matter is made up of particles called atoms and molecules

4 0

2 years ago

Read 2 more answers

What combines to form compounds? A. atoms of elements B. molecules C. minerals D. valences

hammer [34]

D would be correct (i did this Assessment today)<span />

8 0

2 years ago

Read 2 more answers

Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and

notka56 [123]

Answer:

$pH=4.56$

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

$n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol$