The complete reaction of the problem, for better illustration, is
FeO(s) + CO(g) <--> Fe(s) + CO2(g)
The double-tailed arrow signifies that the reaction is in a dynamic chemical equilibrium. When the system is in equilibrium, the forward and the backward reaction rates have an equal ratio of Kp = 0.403 at 1000°C. The formula for Kp is
Kp = [partial pressure of products]/[partial pressure of reactants]
So, first, let's find the partial pressure of the compounds in the reaction.
FeO(s) + CO(g) <--> Fe(s) + CO2(g)
Initial x 1.58 0 0
Change -1.58 -1.58 +1.58 +1.58
------------------------------------------------------------------
Equilbrium x-1.58 0 1.58 1.58
Kp = [(1.58)(1.58)]/[(x-1.58)] = 0.403
x = 7.77 atm (this is the amount of excess FeO)
Therefore, the partial pressure of CO2 at equilibrium is 1.58 atm. There is no more CO because it has been consumed due to excess FeO.
Answer: The initial temperature was 263 K
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
![\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where,
= initial pressure of gas = 0.939 atm
= final pressure of gas = 1.00 atm
= initial volume of gas = 9.40 L
= final volume of gas = 10.0 L
= initial temperature of gas = ?
= final temperature of gas = ![25^oC=273+25=298K](https://tex.z-dn.net/?f=25%5EoC%3D273%2B25%3D298K)
Now put all the given values in the above equation, we get:
![\frac{0.939\times 9.40}{T_1}=\frac{1.00\times 10.0}{298}](https://tex.z-dn.net/?f=%5Cfrac%7B0.939%5Ctimes%209.40%7D%7BT_1%7D%3D%5Cfrac%7B1.00%5Ctimes%2010.0%7D%7B298%7D)
![T_1=263K](https://tex.z-dn.net/?f=T_1%3D263K)
Thus the initial temperature was 263 K
Answer:
igneous or metamorphic
Explanation:
Those two are sorta relvant
24.4 cm.
<h3>Explanation</h3>
HCl and NH₃ reacts to form NH₄Cl immediately after coming into contact. Where NH₄Cl is found is the place the two gases ran into each other. To figure out where the two gases came into contact, you'll need to know how fast they move relative to each other.
The speed of a HCl or NH₃ molecule depends on its <em>kinetic energy</em>.
![E_\text{k} = 1/2 \; m \cdot v^{2}](https://tex.z-dn.net/?f=E_%5Ctext%7Bk%7D%20%3D%201%2F2%20%5C%3B%20m%20%5Ccdot%20v%5E%7B2%7D)
Where
is the <em>kinetic energy</em> of the molecule,
its mass, and
the square of its speed.
Besides, the <em>kinetic theory</em> <em>of gases</em> suggests that for an ideal gas,
![E_\text{k} \propto T](https://tex.z-dn.net/?f=E_%5Ctext%7Bk%7D%20%5Cpropto%20T)
where
its temperature in degrees kelvins. The two quantities are directly proportional to each other. In other words, the <em>average kinetic energy</em> of molecules shall be the same for <em>any ideal gas </em>at the same<em> temperature</em>. So is the case for HCl and NH₃
![E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3)](https://tex.z-dn.net/?f=E_%5Ctext%7Bk%7D%20%28%5Ctext%7BHCl%7D%29%20%3D%20E_%5Ctext%7Bk%7D%20%28%5Ctext%7BNH%7D_3%29)
![m(\text{HCl}) \cdot v^{2}(\text{HCl}) = E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)](https://tex.z-dn.net/?f=m%28%5Ctext%7BHCl%7D%29%20%5Ccdot%20v%5E%7B2%7D%28%5Ctext%7BHCl%7D%29%20%3D%20E_%5Ctext%7Bk%7D%20%28%5Ctext%7BHCl%7D%29%20%3D%20E_%5Ctext%7Bk%7D%20%28%5Ctext%7BNH%7D_3%29%20%3D%20m%28%5Ctext%7BNH%7D_3%29%20%5Ccdot%20v%5E%7B2%7D%28%5Ctext%7BNH%7D_3%29)
Where
,
, and
the mass, speed, and kinetic energy of an HCl molecule;
,
, and
the mass, speed, and kinetic energy of a NH₃ molecule.
The ratio between the mass of an HCl molecule and a NH₃ molecule equals to the ratio between their <em>molar mass</em>. HCl has a molar mass of 35.45; NH₃ has a molar mass of 17.03. As a result,
. Therefore:
![36.45 /17.03\; m(\text{NH}_3) \cdot v^{2}(\text{HCl}) = m(\text{HCl}) \cdot v^{2}(\text{HCl}) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)](https://tex.z-dn.net/?f=36.45%20%2F17.03%5C%3B%20m%28%5Ctext%7BNH%7D_3%29%20%5Ccdot%20v%5E%7B2%7D%28%5Ctext%7BHCl%7D%29%20%3D%20m%28%5Ctext%7BHCl%7D%29%20%5Ccdot%20v%5E%7B2%7D%28%5Ctext%7BHCl%7D%29%20%3D%20m%28%5Ctext%7BNH%7D_3%29%20%5Ccdot%20v%5E%7B2%7D%28%5Ctext%7BNH%7D_3%29)
![36.45 /17.03\; v^{2}(\text{HCl}) = v^{2}(\text{NH}_3)](https://tex.z-dn.net/?f=36.45%20%2F17.03%5C%3B%20v%5E%7B2%7D%28%5Ctext%7BHCl%7D%29%20%3D%20v%5E%7B2%7D%28%5Ctext%7BNH%7D_3%29)
![\sqrt{36.45 /17.03}\; v(\text{HCl}) = v(\text{NH}_3)](https://tex.z-dn.net/?f=%5Csqrt%7B36.45%20%2F17.03%7D%5C%3B%20v%28%5Ctext%7BHCl%7D%29%20%3D%20v%28%5Ctext%7BNH%7D_3%29)
The <em>average </em>speed NH₃ molecules would be
<em>if</em> the <em>average </em>speed of HCl molecules
is 1.
![\text{Time before the two gases meet} = \frac{\text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}](https://tex.z-dn.net/?f=%5Ctext%7BTime%20before%20the%20two%20gases%20meet%7D%20%3D%20%5Cfrac%7B%5Ctext%7BLength%20of%20the%20Tube%7D%7D%7Bv%28%5Ctext%7BHCl%7D%29%20%2B%20v%28%5Ctext%7BNH%7D_3%29%7D)
![\text{Distance from the HCl end} = v(\text{HCl}) \times \text{Time before the two gases meet}\\\phantom{\text{Distance from the HCl end}} = v(\text{HCl}) \times \frac{ \text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}\\\phantom{\text{Distance from the HCl end}} = \frac{v(\text{HCl})}{v(\text{HCl}) + v(\text{NH}_3)} \times \text{Length of the Tube}\\\phantom{\text{Distance from the HCl end}} = \frac{1}{1 + 1.463} \times 60.0\; \text{cm} \\\phantom{\text{Distance from the HCl end}} = 24.4 \; \text{cm}](https://tex.z-dn.net/?f=%5Ctext%7BDistance%20from%20the%20HCl%20end%7D%20%3D%20v%28%5Ctext%7BHCl%7D%29%20%5Ctimes%20%5Ctext%7BTime%20before%20the%20two%20gases%20meet%7D%5C%5C%5Cphantom%7B%5Ctext%7BDistance%20from%20the%20HCl%20end%7D%7D%20%3D%20v%28%5Ctext%7BHCl%7D%29%20%5Ctimes%20%5Cfrac%7B%20%5Ctext%7BLength%20of%20the%20Tube%7D%7D%7Bv%28%5Ctext%7BHCl%7D%29%20%2B%20v%28%5Ctext%7BNH%7D_3%29%7D%5C%5C%5Cphantom%7B%5Ctext%7BDistance%20from%20the%20HCl%20end%7D%7D%20%3D%20%5Cfrac%7Bv%28%5Ctext%7BHCl%7D%29%7D%7Bv%28%5Ctext%7BHCl%7D%29%20%2B%20v%28%5Ctext%7BNH%7D_3%29%7D%20%5Ctimes%20%5Ctext%7BLength%20of%20the%20Tube%7D%5C%5C%5Cphantom%7B%5Ctext%7BDistance%20from%20the%20HCl%20end%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B1%20%2B%201.463%7D%20%5Ctimes%2060.0%5C%3B%20%5Ctext%7Bcm%7D%20%5C%5C%5Cphantom%7B%5Ctext%7BDistance%20from%20the%20HCl%20end%7D%7D%20%3D%2024.4%20%5C%3B%20%5Ctext%7Bcm%7D)
Answer:
B. CH3CH2OH
Explanation:
Ethanol has a chemical formula of CH3CH2OH, it is the second member of the series in the alkanol family. Ethanol is a colourless, volatile liquid with a characteristic smell and taste. It is readily soluble in water in all proportions. It has a boiling point of 78° C. The physical properties such as the solubility of alkanols are affected by the presence of hydrogen bonding. The hydroxyl group is capable of bonding to other alkanol molecules. The boiling points rise with increasing molecular mass.
Hydrogen bonding helps the molecules to stick together. For example comparing the boiling point of pentane ( 36° C) with that of butan-1-ol (118° C) , the boiling point of alkanol is much higher even though the two compound are of similar relative molecular mass. This is due to the presence of hydrogen bonds in butanol.
Hydrocarbons are not soluble in water but alkanols are soluble in water because of the hydroxyl groups in the molecules can form hydrogen bond with water. Solubility of alkanol in water decreases as the number of carbon atom increases. Primary alcohol with more than five carbon atoms are insoluble in water.