Answer:
n(HCl)=1.96 mol
Explanation:
CH4+4Cl2⟶CCl4+4HCl
CCl4+2HF⟶CCl2F2+2HCl
With ideal yields we will end up with 4 moles of HCl.
With 70% yields on every stage
n(HCl)=0.7*0.7*4=1.96 mol
2.1648 kg of CH4 will generate 119341 KJ of energy.
Explanation:
Write down the values given in the question
CH4(g) +2 O2 → CO2(g) +2 H20 (g)
ΔH1 = - 802 kJ
2 H2O(g)→2 H2O(I)
ΔH2= -88 kJ
The overall chemical reaction is
CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ
CH4 +2 O2 → CO2 +2 H20
(1mol)+(2mol)→(1mol+2mol)
Methane (CH4) = 16 gm/mol
oxygen (O2) =32 gm/mol
Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O
which generate 882 KJ /mol
Therefore to produce 119341 KJ of energy
119341/882 = 135.3 mol
to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require
=135.3 *16
=2164.8 gm
=2.1648 kg of CH4
2.1648 kg of CH4 will generate 119341 KJ of energy
Two changes would make this reaction reactant-favored
C. Increasing the temperature
D. Reducing the pressure
<h3>Further explanation</h3>
Given
Reaction
2H₂ + O₂ ⇒ 2H₂0 + energy
Required
Two changes would make this reaction reactant-favored
Solution
The formation of H₂O is an exothermic reaction (releases heat)
If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)
While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient
in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2
As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)
And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater
[H_{3}O^{+}] = 0.00770 M
The equilibrium equation representing the dissociation of 
Given [H_{3}O^{+}] = 0.00770 M
Let the initial concentration of acid be x and change y
So y = ![[H_{3}O^{+}]](https://tex.z-dn.net/?f=%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%20%20)
=![[FCH_{2}COO^{-}]](https://tex.z-dn.net/?f=%20%5BFCH_%7B2%7DCOO%5E%7B-%7D%5D%20%20)
= 0.00770 M
0.00257 x - 0.00001979 = 0.00005929
x = 0.031 M
Therefore, initial concentration of the weak acid is <u>0.031 M</u>
Answer:
q= 110.5 ke
Explanation:
Dipole moment is the product of the separation of the ends of a dipole and the magnitude of the charges.
μ = q * d
μ= Dipole moment (1.93 D)
q= partial charge on each pole
d= separation between the poles(109 pm).
e= electronic charge ( 1.60217662 × 10⁻¹⁹ coulombs)
So,
q= 
coulombs
q = 
e
q = 1.105 * 10⁵ e
q= 110.5 ke
