Density of the gas is 3.05 × 10⁻³ g / cm³.
<u>Explanation:</u>
Volume of the cylinder = π r² h
where r is the radius and h is the height of the height or the length of the glass tube.
Here r = 4 cm and h = 27.4 cm
Volume of the cylinder = 3.14 × 4 × 4 × 27.4 = 1376.6 cm³
We have to find the mass of the gas by subtracting the mass of the tube filled with the substance from the mass of the empty tube.
Mass of the substance = 258.5 - 254.3 = 4.2 g
We have to find the density using the formula as,
Plugin the values as,
= 3.05 × 10⁻³ g / cm³
So the Density of the gas is 3.05 × 10⁻³ g / cm³.
Answer: It changed in identity and properties.
Explanation:
The answer is a.n=1 because it makes sence
Neutralization reactions are the reactions type which form salts.
Explanation:
Salts are formed by ionic bonds when the oxidation states of anions and cations are equal and have opposite signs. So one should be highly electronegative in nature and another should be highly electropositive in nature. So the electropositive element will be ready to give electrons and the electronegative element will be ready to accept all the electrons given by the electropositive element. As a whole the compound will be neutrally charged by adding of equal number of positively charged and negatively charged ions.
The reduction or addition of electrons will be occurring in cations and the oxidation or removal of electrons will be occurring in anions.
So the salt formation is based on neutralization reactions.
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!
