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BigorU [14]
3 years ago
11

Help!!!!!!!!!!!!!!!!!!!

Chemistry
1 answer:
Liono4ka [1.6K]3 years ago
8 0

I think the answer is A.

Hope this helps.

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What is The pH and pOH of a 1.00x10-3M solution of CH3COOH (Ka=1.75x10-5) ?
zaharov [31]

Answer:

<h2>pH = 3.9</h2><h2>pOH = 10.1</h2>

Explanation:

Since CH _ 3COOH is a weak acid to find the pH of CH _ 3COOH we use the formula

pH =  -  \frac{1}{2}   log(Ka)  -  \frac{1}{2}  log(c)

where

Ka is the acid dissociation constant

c is the concentration

From the question

Ka of CH _ 3COOH = 1.75 × 10^-5

c = 1.00 × 10-³M

Substitute the values into the above formula and solve for the pH

That's

pH =  \frac{1}{2} ( -  log(1.75 \times {10 }^{ - 5} -  log(1.00 \times  {10}^{ - 3} )  )  \\   =  \frac{1}{2} (4.757  + 3) \\  =  \frac{1}{2}  \times 7.757) \\  = 3.8785 \:  \:  \:  \:  \:  \:  \:  \:

We have the answer as

<h3>pH = 3.9</h3>

To find the pOH we use the formula

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 3.9

We have the answer as

<h3>pOH = 10.1</h3>

Hope this helps you

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3 years ago
Why doesn't light get absorbed
ziro4ka [17]
It isn't a materialistic object.

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3 years ago
CH2OHCH2OH is a general example of: an ethyl alcohol a methyl alcohol a polyhydroxyl alcohol an organic acid
RoseWind [281]

CH2OHCH2OH is a general example of a polyhydroxyl alcohol. A polyhydroxyl alchol is one in which there are two hydroxyl groups present in the substance. The –OH group attached to both the carbon atoms.

4 0
3 years ago
Read 2 more answers
At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +
Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

4 0
2 years ago
List three properties of metals that nonmetals typically do not have
Law Incorporation [45]
Conductivity, malleability, and high melting points. Hope this helps :)
7 0
3 years ago
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