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pishuonlain [190]

3 years ago

5

How do scientific models help us teach students about systems?

1 answer:

kotegsom [21]3 years ago

4 0

D)By increasing or decreasing the size of systems that are difficult to study we make it easier for students to see how they work and therefore make it easier for them to learn.

Explanation:

Scientific models makes it easier to teach students about systems because their sizes can be adjusted and this makes it easier for students to see how they work.

A model is a simplification of the real work. It takes a part of the real world and studies it.

Models are highly desired in teaching and understanding very complex systems.

Since a tutor owns the control of the model, they can make adjustments on them to a scale that is convenient to work with.
Also, a part of a system can be studied one at a time making it simple for students to comprehend.

learn more:

Models brainly.com/question/1858613

#learnwithBrainly

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Usimov [2.4K]

A. Strong & Weak nuclear forces which are attractive. And electromagnetic.
b. Because the more electronegative atom really wants to complete it's valence shell, so it either covalently, or non-covalently bonds to the other atom.
c. Ummm, ask google? Well, it's kind of logical as well, but the part that Coulomb's law plays into it - I do not know.

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Drawing a conclusion derived from knowledge of an established principle is called deductive reasoning.

allsm [11]

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3 years ago

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What salt would form when each of the following

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A.KCl
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4 0

3 years ago

The molar solubilities of the following compounds (in mol/L) are:

IceJOKER [234]

<u>Answer:</u> The decreasing order of $K_{sp}$ is $AgSCN>AgBr>AgCN$

<u>Explanation:</u>

<u>For AgBr:</u>

The balanced equilibrium reaction for the ionization of silver bromide follows:

$AgBr\rightleftharpoons Ag^{+}+Br^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][Br^-]$

We are given:

$s=7.3\times 10^{-7}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}$

Solubility product of AgBr = $5.33\times 10^{-13}$

<u>For AgCN:</u>

The balanced equilibrium reaction for the ionization of silver cyanide follows:

$AgCN\rightleftharpoons Ag^{+}+CN^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][CN^-]$

We are given:

$s=7.7\times 10^{-9}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}$

Solubility product of AgCN = $5.33\times 10^{-17}$

<u>For AgSCN:</u>

The balanced equilibrium reaction for the ionization of silver thiocyanate follows:

$AgSCN\rightleftharpoons Ag^{+}+SCN^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][SCN^-]$

We are given:

$s=1.0\times 10^{-6}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}$

Solubility product of AgSCN = $1.0\times 10^{-12}$

The decreasing order of $K_{sp}$ follows:

$AgSCN>AgBr>AgCN$

4 0

4 years ago