Answer:
10−8 M.
Explanation:
In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M.
<u>Given:</u>
The initial energy of the electron Einitial = 16.32 * 10⁻¹⁹ J
The energy released i.e the change in energy ΔE = 5.4 * 10⁻¹⁹ J
<u>To determine:</u>
The final energy state Efinal of the electron
<u>Explanation:</u>
Since energy is being released, this suggests that Efinal < Einitial
i.e. ΔE = Einitial - Efinal
Efinal = Einitial - ΔE = (16.32 - 5.4)*10⁻¹⁹ = 10.92 * 10⁻¹⁹ J
Ans: A)
The electron moved down to an energy level and has an energy of 10.92 * 10⁻¹⁹ J
Use the clapeyron equation:
T in kelvin : 6.80 + 273 => 279.8 K
R = 0.082
n = 71.5 moles
P = 5.03 atm
Therefore:
P x V = n x R x T
5.03 x V = 71.5 x 0.082 x 279.8
5.03 x V = 1640.4674
V = 1640.4674 / 5.03
V = 326.13 L
hope ths helps!
The answer to your question is green
Answer: Option (b) and (d) are the correct answer.
Explanation:
Kinetic products are defined as the products which contain a terminal double bond and the reaction is irreversible in nature.
Kinetic controlled products are formed faster because these tend to lower the activation energy. Due to this molecules with less energy are also able to participate in the reaction.
Therefore, rate of reaction increases leading to rapid formation of products.
Therefore, we can conclude that the products of a reaction under kinetic control are product that is formed at the fastest rate and product whose formation requires the smallest free energy of activation.
