Answer:
5
Explanation:
Firstly, we convert what we have to percentage compositions.
There are two parts in the molecule, the sulphate part and the water part.
The percentage compositions is as follows:
Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%
The water part = 100 - 64 = 36%
Now, we divide the percentages by the molar masses.
For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol
For the H2O = 2(1) + 16 = 18g/mol
Now we divide the percentages by these masses
Sulphate = 64/160 = 0.4
Water = 36/18 = 2
The ratio is thus 0.4:2 = 1:5
Hence, there are 5 water molecules.
Answer:
see explanation below
Explanation:
First to all, this is a redox reaction, and the reaction taking place is the following:
2KMnO4 + 3H2SO4 + 5H2O2 -----> 2MnSO4 + K2SO4 + 8H2O + 5O2
According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:
MhVh = MpVp
h would be the hydrogen peroxide, and p the permanganate.
But like it was stated before, the mole ratio is 5:2 so:
5MhVh = 2MpVp
Replacing moles:
5nh = 2MpVp
Now, we just have to replace the given data:
nh = 2MpVp/5
nh = 2 * 1.68 * 0.0213 / 5
nh = 0.0143 moles
Now to get the mass, we just need the molecular mass of the peroxide:
MM = 2*1 + 2*16 = 34 g/mol
Finally the mass:
m = 0.0143 * 34
m = 0.4862 g
<u>Answer:</u> The standard free energy change of formation of 
is 92.094 kJ/mol
<u>Explanation:</u>
We are given:
Relation between standard Gibbs free energy and equilibrium constant follows:
where,
= standard Gibbs free energy = ?
R = Gas constant = 
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
K = equilibrium constant or solubility product = 
Putting values in above equation, we get:
For the given chemical equation:
The equation used to calculate Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the Gibbs free energy change of the above reaction is:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag%5E%2B%28aq.%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag_2S%28s%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol](https://tex.z-dn.net/?f=285.794%3D%5B%282%5Ctimes%2077.1%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%28-39.5%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%3D92.094J%2Fmol)
Hence, the standard free energy change of formation of is 92.094 kJ/mol
False because that doesn’t make sense
Answer:
201.9
Explanation:
when you divided 760 with 400 yo get 19.0 the add it with 200 you get that answer
