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masha68 [24]
3 years ago
15

Scientists often investigate alternative explanations for the data and observations presented, even once a hypothesis has been t

ested and supported through an experiment. This investigation of alternative explanations ...
Chemistry
2 answers:
rodikova [14]3 years ago
8 0

Answer:

Strengthens the evidence and support for a scientific theory.

Explanation:

Hello,

Scientific method provides a compelling tool scientists use to both develop and demonstrate new theories. As it involves both the observation and experimentation towards a specific subject of matter, it turns out convenient to consider alternative explanations substantiating such subject of matter in light of obtaining a more precise explanation for it. In such a way, this investigation of alternative explanations strengthens the evidence and support for a scientific theory.

Best regards.

dedylja [7]3 years ago
6 0
Strengthens the evidence and support for a scientific theory

This is your answer
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When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of the solution according to th
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The concentration of the sodium chloride would be 0.082 M

<h3>Stoichiometric calculations</h3>

From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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Two<br>moin tunetons of carbohydraus​
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Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
ohaa [14]
<span>Answer: 0.094%


</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x


3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177


4) Solve the equation:


You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:


</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M


5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
8 0
3 years ago
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