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Leno4ka [110]
2 years ago
13

Why do the gases on the periodic table tend to form negative ions?

Chemistry
1 answer:
boyakko [2]2 years ago
8 0
Because the valence shell of gases wants to become full
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A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as
umka2103 [35]

Answer:

I would expect to extract the acetic acid.

Explanation:

In the first step, since we are adding a concentrated acid,<u> it will react with the bases present in the mixture (diethylamine and ammonia) </u><u>forming salts</u><u>, </u><u>which are soluble in water</u>. Therefore, after draining the aqueous layer, we will have phenol and acetic acid left in the organic layer.

In the second step, we are adding a diluted base, so it will react with a strong acid. This compound is acetic acid, and its salt will be present in the aqueous layer. Phenol will be left on the organic layer.

7 0
2 years ago
Please help *chemistry*
Daniel [21]
3 because I just did it
8 0
3 years ago
A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2
Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL

Putting values in above equation, we get:

2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL

M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M

0.823 M was the molarity of the KOH solution.

7 0
3 years ago
Indique el valor de los números n,l,m y s para el último electrón de ₇N.(1 pt c/u) n = _____ l = _____ m = _____ s =_____
tia_tia [17]

Answer:

n = 2

l = 1

m = 1

s = +1/2

Explanation:

₇N tiene la configuración electrónica;

1s2 2s2 2p3

Esto implica que este último electrón tiene los siguientes números cuánticos;

n = 2

l = 1

m = 1

s = +1/2

Este último electrón estará en un orbital de 2pz como lo muestran los números cuánticos enumerados anteriormente.

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