Answer: 
Explanation:
Electron gain enthalpy is defined as energy released on addition of electron to an isolated gaseous atom.
The amount of energy released will be maximum when the tendency to attract electrons is maximum. As flourine has atomic number of 9 and has electronic configuration of 2,7. It can readily gain 1 electron to attain stable noble gas configuration and hence liberates maximum energy.
Answer:
D. The equipment needed to accommodate the high temperature and pressure will be expensive to produce.
Explanation:
Hello!
In this case, for the considered reaction, it is clear it is an exothermic reaction because it produces energy; and therefore, the higher the temperature the more reactants are yielded as the reverse reaction is favored. Moreover, since the effect of pressure is verified as favoring the side with fewer moles; in this case the products side (2 moles of ammonia).
In such a way, the high pressure favors the formation of ammonia whereas the high temperature the formation of hydrogen and nitrogen and therefore, option A is ruled out. Since the high pressure shifts the reaction rightwards and the high temperature leftwards, we would not be able to know whether the reaction has ended or not because it will be a "go and come back" process, that is why B is also discarded. Now, since hydrogen and nitrogen would be the "wastes", we discard C because they are not toxic. That is why the most accurate answer would be D. because it is actually true that such equipment is quite expensive.
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The cost of gasoline for the trip would cost $57.28 because you first divide 320 by 20 which gives you 16 than after that you do 16 multiplied by 3.58 which gives you $57.28 as your final product
Answer:
Ka = 
Explanation:
Initial concentration of weak acid =
pH = 6.87
![pH = -log[H^+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%2B%5D)
![[H^+]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D)
![[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-6.87%7D%3D1.35%20%5Ctimes%2010%5E%7B-7%7D%5C%20M)
HA dissociated as:

(0.00045 - x) x x
[HA] at equilibrium = (0.00045 - x) M
x = 
![Ka = \frac{[H^+][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)

0.000000135 <<< 0.00045

Answer:
The answer to your question is 0.64 M
Explanation:
Data
Sucrose C₁₂H₂₂O₁₁ mass = 7.633 g
volume = 25 ml
Molarity = ?
Process
1.- Calculate the molar weight of Sucrose
C₁₂H₂₂O₁₁ = (12 x 12) +(22 x 1) + (11 x 16)
= 144 + 22 + 176
= 342 g
2.- Calculate the moles of sucrose
342 g ------------------ 1 mol
7.633 g --------------- x
x = (7.633 x 1) / 342
x = 0.0223 moles
3.- Calculate the molarity
Molarity = moles / volume (L)
Molarity = 0.0223 / 0.035
Molarity = 0.64