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Oliga [24]
2 years ago
9

How many sublevels are in L energy level?

Chemistry
2 answers:
nlexa [21]2 years ago
7 0

Answer:

14

Explanation:

OLga [1]2 years ago
3 0
The awnser for this is 14
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Salt formed by the partial neutralization of an acid by a base are
Makovka662 [10]

Answer:

D. Acidic salt

Explanation:

Acidic salts:

Salts formed by incomplete neutralisation of poly-basic acids are called acidic salts. Such salts still contain one or more replaceable hydrogen atoms. These salts when neutralised by bases form normal salts.

4 0
2 years ago
Help please Thank You so much!
Irina18 [472]
If you count the number of electrons (small dots), you have the atomic number. In this case you have 11 so this atom is a sodium atom. Sodium has 1 valence electron (electron on the outer shell) and chlorine has 7. This means that if sodium gave one electron away and chlorine would obtain one electron, they would both have the (ideal) noble gas conformation (full outer shell).
4 0
3 years ago
If you have 3.0 moles of argon gas at STP, how much volume will the argon take up?
Anestetic [448]
If you have 3.0 moles of argon gas at STP u would take up 2.5 volume
3 0
3 years ago
Determine all values of hh and k for which the system has no solution
solniwko [45]
Explain more so i could answer it!!
6 0
2 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
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