Answer : 2.31 moles of Helium was initially present in the balloon.
Explanation : Calculating the initial number of moles of helium in the balloon we need to use this method,
final moles present - 14.6 mol
Volume increased from 358.5 mL (= 0.3585 L ) to 2.26 L
<span>(14.6 mol) x (0.3585 L / 2.26 L) = 14.6 moles X 0.158 L = 2.31 moles / L</span>
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<span>Hence, the initial moles was 2.31 moles/L in the helium balloon. </span>
Answer:
![Y=30.6\%](https://tex.z-dn.net/?f=Y%3D30.6%5C%25)
Explanation:
Hello,
In this case, given the reaction, the molar mass of ethene is 28 g/mol and the molar mass of carbon dioxide is 44 g/mol. With that information we compute the theoretical yield considering a 1:2 molar ratio respectively between them:
![m_{CO_20}^{theoretical}=170.9gC_2H_4*\frac{1molC_2H_4}{28gC_2H_4}*\frac{2molCO_2}{1molC_2H_4} *\frac{44gCO_2}{1molCO_2} =537.1gCO_2](https://tex.z-dn.net/?f=m_%7BCO_20%7D%5E%7Btheoretical%7D%3D170.9gC_2H_4%2A%5Cfrac%7B1molC_2H_4%7D%7B28gC_2H_4%7D%2A%5Cfrac%7B2molCO_2%7D%7B1molC_2H_4%7D%20%2A%5Cfrac%7B44gCO_2%7D%7B1molCO_2%7D%20%3D537.1gCO_2)
Thus, we compute the percent yield with the given grams of carbon dioxide:
![Y=\frac{m_{CO_2}^{real}}{m_{CO_2}^{theoretical}}*100 \% =\frac{164.1g}{537.1gCO_2} *100 \%\\\\Y=30.6\%](https://tex.z-dn.net/?f=Y%3D%5Cfrac%7Bm_%7BCO_2%7D%5E%7Breal%7D%7D%7Bm_%7BCO_2%7D%5E%7Btheoretical%7D%7D%2A100%20%5C%25%20%3D%5Cfrac%7B164.1g%7D%7B537.1gCO_2%7D%20%2A100%20%5C%25%5C%5C%5C%5CY%3D30.6%5C%25)
Regards.
Answer:
Below.
Explanation:
Because of the hydrogen bonding that occurs in water. Oxygen is more electronegative than sulphur and forms inter molecular bonding between the H+ and O- ions in the water. This does not occur in H2S.
11.
c, because the object is not in direct contact to anything with gravity