Answer:
water in the flat plate is less so it takes lesser time to disappear
Explanation:
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)
So for D you have to find energy right
from c you get wavelength Lambda
so on D use this
E = Hc / lambda
c is given 1.5 x 10 ^20
h = 6.624 x 10^-34
and then you get answer for energy
Answer:
Conservation of mass mean mass cannot be created nor destroyed
Explanation:
<u>Answer:</u> The standard Gibbs free energy of the reaction is -15.8 kJ/mol
<u>Explanation:</u>
Relation between standard Gibbs free energy and equilibrium constant follows:
![\Delta G^o=-RT\ln K_{eq}](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Cln%20K_%7Beq%7D)
where,
= Standard Gibbs free energy = ?
R = Gas constant = ![8.314J/K mol](https://tex.z-dn.net/?f=8.314J%2FK%20mol)
T = temperature = 298 K
= equilibrium constant = 581
Putting values in above equation, we get:
![\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (581)\\\\\Delta G^o=-15769.13J/mol=-15.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-%288.314J%2FKmol%29%5Ctimes%20298K%5Ctimes%20%5Cln%20%28581%29%5C%5C%5C%5C%5CDelta%20G%5Eo%3D-15769.13J%2Fmol%3D-15.8kJ%2Fmol)
Conversion factor used: 1 kJ = 1000 J
Hence, the standard Gibbs free energy of the reaction is -15.8 kJ/mol