Solutions are homogeneous mixtures made up of two non-reacting compounds. The component which is present in large quantity is called as solvent and the component in small quantity is called as solute There are six types of solutions.
1. Gas (solvent) - Gas (solute) Example= Air
2. Liquid (solvent) - Gas (solute) Example= Fizzy Drinks
3. Liquid (solvent) - Liquid (solute) Example= 96% Ethanol in Water
4. Liquid (solvent) - Solid (solute) Example= Sugar in Tea
5. Solid (solvent) - Solid (solute) Example= Brass, 18 caret Gold
6. Solid (solvent) - Gas (solute) Example= Marshmellow
Result:
In equation earth air is solvent (as it is in large quantity) and water vapors (gas) which are in small quantity is solute.
So, it is Gas - Gas Solution.
Tritium is an isotope of hydrogen...
For Isotopes number of neutrons are different but electrons and protons are same
Answer:
(d) has an extremely large equilibrium constant.
Explanation:
Hello,
Considering a generic chemical reaction:
![aA+bB cC+dD](https://tex.z-dn.net/?f=aA%2BbB%20%3C--%3E%20cC%2BdD)
The equilibrium constant is defined as:
![K=\frac{[C]^c_{eq}[D]^d_{eq}}{[A]^a_{eq}[B]^b_{eq}} \\](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC%5D%5Ec_%7Beq%7D%5BD%5D%5Ed_%7Beq%7D%7D%7B%5BA%5D%5Ea_%7Beq%7D%5BB%5D%5Eb_%7Beq%7D%7D%20%5C%5C)
Now, an irreversible chemical reaction is a reaction in which the reagents are converted into products with no chance of coming back, so, considering the previous chemical reaction, the concentration of both A and B tends to be zero, so an extremely large equilibrium constant is gotten.
Best regards.
Answer:
NH4Cl, NaCl, Ba(OH)2, NaOH
Explanation:
NH4Cl is an acidic salt formed by the neutralization of a strong acid (HCl) with a weak base (NH3). Hence, it will habe a PH<7 (the lowest PH).
NaCl is a neutral salt,formed by neutralization of a strong acid (HCl) with a strong base (NaOH). Hence, it will have a PH of 7.
Ba(OH)2 is a weak base. Therefore, it will have a PH between 8 and 10.
NaOh meanwhile, is a strong base. Therefore, it will have a PH between 10 to 13.
Hence, we have
NH4Cl < NaCl < Ba(OH)2 < NaOH
Answer:
Equilibrium constant for
is 0.5
Equilibrium constant for decomposition of
is ![1.79 \times 10^{-14}](https://tex.z-dn.net/?f=1.79%20%5Ctimes%2010%5E%7B-14%7D)
Explanation:
dissociates as follows:
![PCl_5 \rightleftharpoons PCl_3+Cl_2](https://tex.z-dn.net/?f=PCl_5%20%5Crightleftharpoons%20PCl_3%2BCl_2)
initial 0.72 mol 0 0
at eq. 0.72 - 0.40 0.40 0.40
Expression for the equilibrium constant is as follows:
![k=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
Substitute the values in the above formula to calculate equilibrium constant as follows:
![k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5B0.40%2F1%5D%5B0.40%2F1%5D%7D%7B0.32%2F1%7D%20%5C%5C%3D%5Cfrac%7B0.40%20%5Ctimes%200.40%7D%7B0.32%7D%20%5C%5C%3D0.5)
Therefore, equilibrium constant for
is 0.5
Now calculate the equilibrium constant for decomposition of ![NO_2](https://tex.z-dn.net/?f=NO_2)
It is given that
is decomposed.
decomposes as follows:
![2NO_2 \rightleftharpoons 2NO + O_2](https://tex.z-dn.net/?f=2NO_2%20%5Crightleftharpoons%202NO%20%2B%20O_2)
initial 1.0 M 0 0
at eq. concentration of
is:
![[NO_2]_{eq}=1-(0.000066) = 0.999934\ M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D1-%280.000066%29%20%3D%200.999934%5C%20M)
![[NO]_{eq}=6.6 \times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D6.6%20%5Ctimes%2010%5E%7B-5%7D%5C%20M)
Expression for equilibrium constant is as follows:
![K=\frac{[NO]^2[O_2]}{[NO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%5BO_2%5D%7D%7B%5BNO_2%5D%5E2%7D)
Substitute the values in the above expression
![K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5B6.6%5Ctimes%2010%5E%7B-5%7D%5D%5E2%5B3.3%20%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5B0.999934%5D%5E2%7D%20%5C%5C%3D1.79%5Ctimes%2010%5E%7B-14%7D)
Equilibrium constant for decomposition of
is ![1.79 \times 10^{-14}](https://tex.z-dn.net/?f=1.79%20%5Ctimes%2010%5E%7B-14%7D)