Answer:
Molarity: 0.522M
Percentage by mass: 2.36 (w/w) %
Explanation:
Formic acid, HCOOH reacts with NaOH as follows:
HCOOH + NaOH → NaCOOH + H₂O
To solve this question we must find the moles of NaOH added = Moles formic acid. Taken into account the dilution that was made we can find the moles -And molarity of formic acid and its percentage by mass as follows:
<em>Moles NaOH = Moles HCOOH:</em>
0.01580L * (0.1322mol / L) =0.002089 moles HCOOH
<em>Moles in the original solution:</em>
0.002089 moles HCOOH * (25mL / 10mL) = 0.005222 moles HCOOH
<em>Molarity of the solution:</em>
0.005222 moles HCOOH / 0.01000L =
<h3>0.522M</h3>
<em>Mass HCOOH in 1L -Molar mass: 46.03g/mol-</em>
0.522moles * (46.03g / mol) = 24.04g HCOOH
<em>Mass solution:</em>
1L = 1000mL * (1.02g / mL) = 1020g solution
<em>Mass percent:</em>
24.04g HCOOH / 1020g solution * 100
2.36 (w/w) %
Answer:
c) NOx and VOC
oxides of nitrogen (NOx) and volatile organic compounds (VOC)
Explanation:
-
The answer is homogeneous
Answer: Gold.
Explanation: Water, CO2, and table salt are compounds. They are composed of two or more separate elements; a mixture, where as gold is neither.
Answer:
Explanation:
Hello!
In this case, since the definition of entropy in a random mixture is:
![\Delta S=-n_TR\Sigma[x_i*ln(x_i)]](https://tex.z-dn.net/?f=%5CDelta%20S%3D-n_TR%5CSigma%5Bx_i%2Aln%28x_i%29%5D)
For this silver-gold mixture we write:
![\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]](https://tex.z-dn.net/?f=%5CDelta%20S%3D-%28n_%7BAu%7D%2Bn_%7BAg%7D%29R%5CSigma%5B%5Cfrac%7Bn_%7BAu%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%2Aln%28%5Cfrac%7Bn_%7BAu%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%29%2B%5Cfrac%7Bn_%7BAg%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%2Aln%28%5Cfrac%7Bn_%7BAg%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%29%5D)
By knowing the moles of gold:
It is possible to write the aforementioned formula in terms of the variable 
representing the moles of silver:
![20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]](https://tex.z-dn.net/?f=20%5Cfrac%7BJ%7D%7Bmol%7D%3D-%280.508%2Bx%298.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%5CSigma%5B%5Cfrac%7B0.508%7D%7B0.508%2Bx%7D%20%2Aln%28%5Cfrac%7B0.508%7D%7B0.508%2Bx%7D%20%29%2B%5Cfrac%7Bx%7D%7B0.508%2Bx%7D%20%2Aln%28%5Cfrac%7Bx%7D%7B0.508%2Bx%7D%20%29%5D)
Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:
So the mass is:
Best regards!
