Answer:
3.65 g of cyclohexene
Explanation:
Cyclohexanol + phosphoric acid ----> cyclohexene
The reaction is 1:1 hence the limiting reactant is phosphoric acid.
Hence,
1 mole of phosphoric acid yields 1 mole of cyclohexene
0.0444 moles of phosphoric acid yields 0.0444 moles of cyclohexene
Theoretical yield = number of moles of cyclohexene × molar mass of cyclohexene
Theoretical yield = 0.0444 moles of cyclohexene × 82.143 g/mol
Theoretical yield = 3.65 g of cyclohexene
The reaction N2O4 (g) <--> 2NO2 (g) is endothermic, meaning that it consumes heat to move towards formation of the products.
According to Le Chatelier's Principle, therefore, if heat is added, more product (NO2) will be produced, and equilibrium would shift towards the right side. This is choice 3.
Answer:
1. 
2. 
Explanation:
Hello,
1. In this case, since the volume of the rock is obtained via the difference between the volume of the cylinder with the water and the rock and the volume of the cylinder with the water only:

Thus, the density turns out:

2. In this case, given the density and mass of aluminum we can compute its volume as follows:

Moreover, as the volume is also defined in terms of width, height and length:

The height is computed to be:

Best regards.
Basically all of the elements found in Group I of the periodic table also have this property. The ability to easily give up a single valence electron.
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is 