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12345 [234]
3 years ago
10

An alum hydrate sample was analyzed by decomposition and gave the following data:

Chemistry
1 answer:
xz_007 [3.2K]3 years ago
7 0

Answer:

No, the experimental result is different from the theoretical value.

Explanation:

Based on the given information, the mass of beaker and watchglass plus alum hydrate is 102.218 grams, and the mass of beaker and watchglass is 101.286 grams. Therefore, the mass of alum hydrate is:  

= 102.218 grams - 101.286 grams

= 0.932 grams

Now the mass of anhydrous compound is,  

= 102.218 grams - 101.798 grams

= 0.42 grams

Thus, the mass of water present is,  

= 0.932 grams - 0.42 grams

= 0.512 grams

The mass percent of water is,  

= mass of water/Total mass of hydrate * 100

= 0.512 grams / 0.932 grams * 100

= 54.93 %

Hence, the experimental result in not similar to the theoretical result.  

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The reaction takes place in water. What happens to the equilibrium when the pressure is increased? a)It favors formation of prod
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I don't really get the options but it favoures the reactant side.

Explanation:

Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g

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With the explanations I have made, I hope the question is now clear to you.

6 0
4 years ago
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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