Given reactions:
(A) 6CO2(g) + 6H2O(l) + sunlight → C6H12O6(aq) + 6O2(g)
(B) 2H2(g) + O2(g) → 2H2O(g) + energy
Exothermic reactions are those which proceed with the release of heat/energy. In contrast, endothermic reactions proceed with the absorption of energy in the form of heat or light.
Since reaction A required sunlight, it is endothermic. Reaction B releases energy, hence exothermic
Ans: (B)
A is endothermic
B is exothermic
Answer:
only chlorine can expand its octet.
Explanation:
An atom can expand its octet is it has empty d orbital
the electronic configuration of given elements will be:
B : 1s2 2s2 2p1 [Valence shell n =2 no d orbital]
O :1s2 2s2 2p4 [Valence shell n =2 no d orbital]
F : 1s2 2s2 2p5 [Valence shell n =2 no d orbital]
Cl :1s2 2s2 2p6 3s2 3p5 3d0 [Valence shell n =2 no d orbital]
Out of given elements only chlorine has empty d orbitals in its valence shell
Thus only chlorine can expand its octet.
Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:
Here, 
is mole fraction of A, 
is mole fraction of B, 
is partial pressure of A and 
is partial pressure of B.
The mole fraction of A and B are related to each other as follows:
In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:
Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.
Now, vapor pressure can be calculated as follows:
Putting the values,
Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.
Answer:
Check the explanation
Explanation:
When,
pH = -log[H+] = 3.30
[H+] = 
![alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6](https://tex.z-dn.net/?f=alpha%5BY%5E-4%5D%20%3D%20%5BH%2B%5D%5E6%20%2B%20Ka1%5BH%2B%5D%5E5%20%2B%20Ka1Ka2%5BH%2B%5D%5E4%20%2B%20Ka1Ka2Ka3%5BH%2B%5D%5E3%20%2B%20Ka1Ka2Ka3Ka4%5BH%2B%5D%5E2%20%2B%20Ka1Ka2Ka3Ka4Ka5%5BH%2B%5D%20%2B%20Ka1Ka2Ka3Ka4Ka5Ka6)
= 
= 
When,
pH = -log[H+] = 10.15
[H+] = 
Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 
; Ka6 = 
![alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6](https://tex.z-dn.net/?f=alpha%5BY%5E%7B-4%7D%5D%20%3D%20%5BH%2B%5D%5E6%20%2B%20Ka1%5BH%2B%5D%5E5%20%2B%20Ka1Ka2%5BH%2B%5D%5E4%20%2B%20Ka1Ka2Ka3%5BH%2B%5D%5E3%20%2B%20Ka1Ka2Ka3Ka4%5BH%2B%5D%5E2%20%2B%20Ka1Ka2Ka3Ka4Ka5%5BH%2B%5D%20%2B%20Ka1Ka2Ka3Ka4Ka5Ka6)
= 
= 
The moles of any substance are equal to the substance's mass divided by its molar mass. Therefore, in order to calculate the moles of copper, you would divide the reacted mass by 63.55
