Well, Tin(Sn) is in Period 5 and a member of group 14.
Molar mass of FE2O3=2(55.85)+3(16)=159.7
2.56g*1mol/159.7*2mol/1mol*55.85g/1mol=1.79g
Answer: one reason may be that since the lake is warmer, it may contain lower levels of oxygen.
Explanation: Temperature Is just one factor (along with pressure) that dictates the amount of gas that can be dissolved in a liquid.
Answer:
The unbalanced chemical equation of propane combustion is
C₃H₈ (g) + O₂ (g) → CO₂(g) + H₂O(g)
Explanation:
Combustion is a type of chemical reaction in which there is the release of energy in the form of heat. This means that this reaction can be called exothermic.
When propane reacts with oxygen present in the air, it releases a high amount of heat. If the reaction is fully completed, the products are carbon dioxide and water vapor.
Answer:
![n_{O_2}=0.866molSO_2\\\\n_{SO_2}=0.577molSO_2](https://tex.z-dn.net/?f=n_%7BO_2%7D%3D0.866molSO_2%5C%5C%5C%5Cn_%7BSO_2%7D%3D0.577molSO_2)
Explanation:
Hello there!
In this case, according to the given balanced chemical reaction by which ZnS reacts with O2, it is possible to calculate the moles of the latter that are consumed, not produced, according the 2:3 mole ratio between them and the following stoichiometric set up:
![n_{O_2}=56.25gZnS*\frac{1molZnS}{97.47gZnS}*\frac{3molO_2}{2molZnS} \\\\n_{O_2}=0.866molO_2](https://tex.z-dn.net/?f=n_%7BO_2%7D%3D56.25gZnS%2A%5Cfrac%7B1molZnS%7D%7B97.47gZnS%7D%2A%5Cfrac%7B3molO_2%7D%7B2molZnS%7D%20%20%5C%5C%5C%5Cn_%7BO_2%7D%3D0.866molO_2)
But also, we can compute the moles of SO2 that are produced via the the 2:2 mole ratio of ZnS to SO2:
![n_{SO_2}=56.25gZnS*\frac{1molZnS}{97.47gZnS}*\frac{2molSO_2}{2molZnS} \\\\n_{SO_2}=0.577molSO_2](https://tex.z-dn.net/?f=n_%7BSO_2%7D%3D56.25gZnS%2A%5Cfrac%7B1molZnS%7D%7B97.47gZnS%7D%2A%5Cfrac%7B2molSO_2%7D%7B2molZnS%7D%20%20%5C%5C%5C%5Cn_%7BSO_2%7D%3D0.577molSO_2)
Regards!