Argon has 24 known isotopes.
Always before starting any calculation it is important to verify that the reaction is balanced. To do this we count the number of atoms of each element on both sides of the reaction. When counting them we have to:
Carbon (C) - 1 atoms
Hydrogen (H) - 4 atoms
Oxygen (O) - 1 atoms
The reaction is balanced, now we proceed to see the stoichiometry of the reaction. We see that for two moles of O2 that react, 1 mole of CO2 is formed, that is, the ratio is 2 to 1.
We must calculate the moles of O2 that correspond to 370.2 grams, for this we use the molar mass of O2 (31.998g/mol).
![\begin{gathered} \text{MolO}_2=GivengO_2\times\frac{1molO_2}{MolasMass,gO_2} \\ \text{MolO}_2=370.2gO_2\times\frac{1molO_2}{31.998gO_2}=11.6molO_2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7BMolO%7D_2%3DGivengO_2%5Ctimes%5Cfrac%7B1molO_2%7D%7BMolasMass%2CgO_2%7D%20%5C%5C%20%5Ctext%7BMolO%7D_2%3D370.2gO_2%5Ctimes%5Cfrac%7B1molO_2%7D%7B31.998gO_2%7D%3D11.6molO_2%20%5Cend%7Bgathered%7D)
Now, the moles of CO2 produced theoretically will be:
![\begin{gathered} \text{MolCO}_2=GivenmolO_2\times\frac{1molCO_2}{2molO_2} \\ \text{MolCO}_2=11.6molO_2\times\frac{1molCO_2}{2molO_2}=5.9molCO_2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7BMolCO%7D_2%3DGivenmolO_2%5Ctimes%5Cfrac%7B1molCO_2%7D%7B2molO_2%7D%20%5C%5C%20%5Ctext%7BMolCO%7D_2%3D11.6molO_2%5Ctimes%5Cfrac%7B1molCO_2%7D%7B2molO_2%7D%3D5.9molCO_2%20%5Cend%7Bgathered%7D)
The grams corresponding to the 5.8 moles of CO2 produced will be the theoretical yield of the reaction, and we find it using the molar mass of CO2 equal to 44.01g/mol:
![\begin{gathered} \text{gCO}_2=GivenmolCO_2\times\frac{MolarMass,gCO_2}{1molCO_2} \\ \text{gCO}_2=5.9molCO_2\times\frac{44.01gCO_2}{1molCO_2}=254.6gCO_2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7BgCO%7D_2%3DGivenmolCO_2%5Ctimes%5Cfrac%7BMolarMass%2CgCO_2%7D%7B1molCO_2%7D%20%5C%5C%20%5Ctext%7BgCO%7D_2%3D5.9molCO_2%5Ctimes%5Cfrac%7B44.01gCO_2%7D%7B1molCO_2%7D%3D254.6gCO_2%20%5Cend%7Bgathered%7D)
The grams of CO2 found to correspond to the theoretical yield, now the percent yield is found with the following equation:
![\text{Percent yield=}\frac{\text{Actual yield}}{Theoretical\text{ yield}}\times100\%](https://tex.z-dn.net/?f=%5Ctext%7BPercent%20yield%3D%7D%5Cfrac%7B%5Ctext%7BActual%20yield%7D%7D%7BTheoretical%5Ctext%7B%20yield%7D%7D%5Ctimes100%5C%25)
Actual yield=85.9g of CO2
Theoretical yield=254.6g of CO2
![\text{Percent yield=}\frac{\text{8}5.9gCO_2}{254.6gCO_2}\times100\%=33.7\%](https://tex.z-dn.net/?f=%5Ctext%7BPercent%20yield%3D%7D%5Cfrac%7B%5Ctext%7B8%7D5.9gCO_2%7D%7B254.6gCO_2%7D%5Ctimes100%5C%25%3D33.7%5C%25)
The percent yield of the reaction is 33.7%
Answers with Explanations:
1. Explain how thermal energy is transferred from the sun to Earth's surface.
"Thermal energy" refers to the<em> energy produced from heat</em>. The sun gives off <em>light energy</em> (solar) through it sunshine. This solar energy penetrates the atmosphere of the earth. Once the atmosphere is heated, it transfers the thermal energy to the Earth's surface in the form of rays. This process is called "radiation."
This can also be done through the process called "conduction." Since air is a poor conductor of heat, thermal energy is only being transferred in the atmosphere. Direct contact between objects is essential in conduction.
It is said that some of the heat transferred to Earth<em> bounces back to the atmosphere.</em> This is because once the Earth's surface heats up,<u> the </u><u>warmest air expands</u><u> and rises towards the atmosphere.</u>
2. How does this energy impact the land and water on Earth?
The sun heats up both the land and water on Earth. However, water is a good conductor of heat so it allows heat to stay in it for a<u> longer period of time. </u>On the contrary, the Earth's land surface is a poor conductor of heat so this allows heat to escape faster than in water. The rays of the sun can penetrate the water at a<em> deeper length</em> than that of the soil.
Answer:
0.3195 mole
Explanation:
Number of moles of C3Hg
= mass ÷ molar mass
= 75.6 g ÷ (12 + 12 + 12 + 200.6)
= 75.6 g ÷ 236.6
= 0.3195 mole