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Nonamiya [84]
3 years ago
13

2. Which one is the modal class?

Chemistry
1 answer:
Sonja [21]3 years ago
7 0

Answer:

a.3-5 is the modal class.

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Calculate the pOH of an aqueous solution of .0.073 M LiOH
Novosadov [1.4K]

Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).

pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.

LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:

[LiOH]= [OH-]= 0.073 M

Replacing in the definition of pOH:

pOH= -log (0.073 M)

<u><em>pOH= 1.14 </em></u>

In summary, the pOH of the aqueous solution is 1.14

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7 0
2 years ago
What two categorical ways of thinking about matter are discussed in the passage?
Mamont248 [21]

Answer:

There's no passage but I can say that matter are that ''thing'' which extend through all Universe, matter is everything. It can be seen or felt as liquid, gas or solid. This changes of its state depend on the amount of energy that is involved.

So, here is used categorical thinking because is a abstract concept that is hard to understand sometimes. It's abstract when we say that matter is all, not  just what we see, but even what we don't see like ''Dark Matter'', which is the majority in the Universe.

4 0
3 years ago
What is oxidized in a galvanic cell with aluminum and gold electrodes ?
k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V

Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Notice that the overall cell potential upon summing is:

E_{cell}=1.66 V + 1.50 V=3.16 V

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0
3 years ago
To make a 2.0-molar solution, how many moles of solute must be dissolved in 0.50 liters of solution?
attashe74 [19]
Moles = Molarity x Volume

Moles = 2.0 x 0.50

= 1.0 mole

hope this helps!

4 0
3 years ago
In one experiment, the reaction of 1.00 mercury and an excess of sulfur yielded 1.16g of a sulfide of mercury
Nuetrik [128]

<u>Answer and Explanation:</u>

Mercury combines with sulfur as follows -

Hg + S = HgS

Hg = 200,59

S = 32,066 Therefore 1.58 g of Hg will react with -

1.58 multiply with 32,066 divide by 200,96 of sulfur.

= 0.25211 g S

This will form 1.58 + 0.25211 g HgS  = 1.83211 g HgS

The amount of S remaining = 1.10 - 0.25211  = 0.84789 g

5 0
3 years ago
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