Answer: CO2x+O4H I don’t know if this is the right answer
Answer:
Option D = No, when elements combine to form a new material, they have properties unique to the new materials.
Explanation:
When sodium contact with water it loses its one electron and thus gain positive charge. When there are more sodium atoms present and many atoms do this thus more positive ions are produced and these positive ions repeal each other at high speed and explosion occur.
But when it form compound with other material, it will not showed this behavior.
Example:
Consider the sodium chloride, when it dissolve in water sodium not showed explosion. In sodium chloride sodium already gives its electron to the chlorine and have stable electronic configuration. The sodium present in cationic form. When it dissolve, partial positive charge of water surrounds the Cl⁻ and partial negative charge of water surrounds the Na⁺ ion, ans sodium chloride gets dissolve into water without explosion.
The chemical formula of Iron (III) Sulfide is FeSO3. This element or compound has another name which is <span>ferric sulfide or sesquisulfide.</span>
Answer:
27%
Explanation:
Hello,
The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:
![2NaOH+H_2SO_4-->Na_2SO_4+2H_2O](https://tex.z-dn.net/?f=2NaOH%2BH_2SO_4--%3ENa_2SO_4%2B2H_2O)
Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:
![n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\](https://tex.z-dn.net/?f=n_%7BNaOH%7D%3D7.8gNaOH%2A%5Cfrac%7B1molNaOH%7D%7B40gNaOH%7D%3D0.2molNaOH%5C%5Cn_%7BH_2SO_4%7D%3D4.9gH_2SO_4%2A%5Cfrac%7B1molH_2SO_4%7D%7B98gH_2SO_4%7D%3D0.050molH_2SO_4%5C%5C)
- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:
![0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4](https://tex.z-dn.net/?f=0.2molNaOH%2A%5Cfrac%7B1molH_2SO_4%7D%7B2molNaOH%7D%3D0.098molH_2SO_4)
As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:
![0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4](https://tex.z-dn.net/?f=0.05molH_2SO_4%2A%5Cfrac%7B1molNa_2SO_4%7D%7B1mol%20H_2SO_4%7D%20%2A%5Cfrac%7B142.04gNa_2SO_4%7D%7B1molNa_2SO_4%7D%20%3D7.1gNa_2SO_4)
Finally, the percent yield turns out into:
![Y=\frac{1.92g}{7.1g} *100](https://tex.z-dn.net/?f=Y%3D%5Cfrac%7B1.92g%7D%7B7.1g%7D%20%2A100)
%
Best regards.