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notka56 [123]
3 years ago
8

A substance in which light can travel through such as air,glass, or water

Chemistry
1 answer:
Stels [109]3 years ago
6 0

Answer:

The correct answer is - transparent medium.

Explanation:

A transparent substance or medium is the substance that allows light to pass through it. Light moves through these substances as they do not absorb the light and do not reflect too.

The example of such substances is glass, air or water. These substances allow light to pass through them.

Thus, The correct answer is - transparent medium.

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Iupac name of h3C-C-Ch2-ch3​
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Answer:

The IUPAC name of given compound is 3−5−ethyl−5−−3−methylheptane. Explanation: The parental chain is of 7 carbons with single bonds hence it is heptane. Two substituents ethyl and methyl group are attached from an equal distance. Hence according to the alphabetical order preference, counting starts from carbon which is close to an ethyl group.

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3 years ago
A mechanical pencil has a mass of 47.4 grams. The volume of the pencil is 15.8 cubic centimeters. What is the density of the pen
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"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"
Salsk061 [2.6K]

Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

Ka=\frac{(0.00195)^2}{0.25-0.00195}

Ka=1.53*10^-^5

So, the value of Ka for butyric acid is 1.53*10^-^5 .

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