Answer:
530.835 g
Explanation:
First we convert 244 g of benzoic acid (C₇H₆O₂) to moles, using its molar mass:
- 244 g benzoic acid ÷ 122.12 g/mol = 2.00 moles benzoic acid
Theoretically,<em> one mol of ethyl benzene would produce one mol of benzoic acid</em>. But the experimental yield tells us that one mol of ethyl benzene will produce only 0.4 moles of benzoic acid.
With the above information in mind we convert 2.00 moles of benzoic acid into moles of ethyl benzene:
- 2.00 moles benzoic acid *
= 5.00 moles ethyl benzene
Finally we <u>convert moles of ethyl benzene </u>(C₈H₁₀)<u> into grams</u>, using its <em>molar mass</em>:
- 5.00 moles ethyl benzene * 106.167 g/mol = 530.835 g ethyl benzene
Answer:
The pressure inside the container would increase with each additional pump.
Explanation:
- From the general gas law of ideal gases:
<em>PV = nRT,</em>
where, P is the pressure of the gas.
V is the volume of the gas.
n is the no. of moles of the gas.
R is the general gas constant.
T is the temperature of the gas.
- As clear from the gas law; the pressure of the gas is directly proportional to the no. of moles of the gas.
<em>P α n.</em>
- As gas particles are pumped into a rigid steel container, the no. of moles of the gas will increase.
So, the pressure of the gas will increase.
<em>Thus, the right choice is: The pressure inside the container would increase with each additional pump.</em>
<u>Answer:</u> The increase in pressure is 0.003 atm
<u>Explanation:</u>
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= final pressure = ?
= Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![801^oC=[801+273]K=1074K](https://tex.z-dn.net/?f=801%5EoC%3D%5B801%2B273%5DK%3D1074K)
= final temperature = ![(801+1.00)^oC=802.00=[802+273]K=1075K](https://tex.z-dn.net/?f=%28801%2B1.00%29%5EoC%3D802.00%3D%5B802%2B273%5DK%3D1075K)
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28800J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B1074%7D-%5Cfrac%7B1%7D%7B1075%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D3%5Ctimes%2010%5E%7B-3%7Datm%5C%5C%5C%5CP_2%3De%5E%7B3%5Ctimes%2010%5E%7B-3%7D%7D%3D1.003atm)
Change in pressure = 
Hence, the increase in pressure is 0.003 atm
thermal energy leads to an increase in the kinetic energy also increasing in temperature hope this helped!
Answer:
84.24 g
Explanation:
Given data:
Mass of oxygen = 75 g
Mass of Al required to react = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 75 g/ 32 g/mol
Number of moles = 2.34 mol
Now we will compare the moles of oxygen with Al.
O₂ : Al
3 : 4
2.34 : 4/3×2.34 = 3.12 mol
Mass of Al required:
Mass = number of moles × molar mass
Mass = 3.12 mol × 27 g/mol
Mass = 84.24 g
