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Musya8 [376]
3 years ago
6

Classify the following as an acid or base using the Arrhenius definition of an acid or base.

Chemistry
2 answers:
Colt1911 [192]3 years ago
6 0
An Arrhenius acid is a substance that releases H+ ions when it dissociates in water; on the other hand, an Arrhenius base releases OH- ions when it dissociates in water. Therefore, the compounds are classified as follows:

1. Acid, H+ ions
2. Base, OH- ions
3. Base, OH- ions
4. Acid, H+ ions
Morgarella [4.7K]3 years ago
6 0

HNO3 is an Arrhenius acid and increases the concentration of H+ when added to water. KOH is an Arrhenius base and increases the concentration of OH- when added to water. Ca(OH) 2 is an Arrhenius acid, and increases the concentration of H+. H2SO4 is an Arrhenius acid, and increases the concentration of H+ when added to water.

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<span>To solve for m in the equation F = ma, you must divide both side of the equation by a. This will make the equation look like F/a = ma/a. Since m is being multiplied by a, dividing it will cancel out. Now making the final equation look like F/a=m and/or m=F/a.</span>
6 0
3 years ago
An emulsifying agent is typically characterized by having
pochemuha

The third substance or agent which produce the film between the interface of two immiscible liquids and thus stabilize the system are known as an emulsifying agent.

Since the solubility of the liquids depends on the polarity of the mixing liquids the thumb rule of solubility is like dissolves like that means polar liquid dissolves in polar liquid only and vice versa. For two immiscible liquids, the emulsifying agent is used which does not chemically change the polarity of liquids but acts as bridge between immiscible liquids, the polar end of the emulsifier attach to the polar liquid and the non-polar end of the emulsifier attach to the non-polar end and thus help in dissolving.

Therefore, the one end of the emulsifier is polar and the other end is non-polar

5 0
4 years ago
If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH
KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)

M_{KOH}=0.113 M\\V_{KOH}=79.06 mL

V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????

M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}

M_{CH3COOH} = \frac{0.113*79.06}{95.27}

M_{CH3COOH} = 0.094 M

Moles of CH_3COOH = 95.27* 10^{-3}* 0.094

=0.0090 moles

Moles of  KOH = 79.06*10^{-3}*0.113

= 0.0090 moles

The equation for the reaction can be expressed as :

CH_3COOH     +      KOH     ----->      CH_3COO^{-}K^+      +     H_2O

Concentration of CH_3COO^{- ion = \frac{0.0090}{Total volume (L)}

= \frac{0.0090}{(95.27+79.06)} *1000

= 0.052 M

Hydrolysis of  CH_3COO^{- ion:

CH_3COO^{-      +       H_2O      ----->      CH_3COOH       +     OH^-

K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}

⇒    \frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}

=     0.5494*10^{-9}= \frac{x*x}{0.052-x}

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

0.5494*10^{-9}= \frac{x^2}{0.052}

x^2 = 0.5494*10^{-9}*0.052

x^2 = 0.286*10^{-10

x = \sqrt{0.286*10^{-10

x =0.535*10^{-5}M

[OH] = x =0.535*10^{-5}

pOH = -log[OH^-]

pOH = -log[0.535*10^{-5}]

pOH = 5.27

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0
3 years ago
Which properties are examples of chemical properties? Check all that apply.
Scorpion4ik [409]

Answer:

1.Reactivity

2.combustibility

3.Shape

Hope this helps,

If you think I am right please give brainliest

Happy Holidays!

Explanation:

5 0
3 years ago
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