Hello!
This is a difficultly easy question, mostly because it seems like all of the answers would be right, but if I had to choose one, I would say (D) because this option gives the most detail.
Option A only tells us how fast a car was moving
Option B only tells us the rate the water falls and how fast it is going
Option C only tells how far and not the speed or destination
Option D is correct, it tells us the speed/rate, the destination, and what is moving
Therefore, Option (D) is correct
If this helps, please mark brainliest, have a great day!!
The sun sets in the west on Earth and most of the planets in our solar system. <span>But the sun doesn't set in the west on Uranus, since Uranus's axis of rotation is tilted by about 90 degrees to the ecliptic.
Hope this helps :)
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Answer:
[OH⁻] = 0.016
Explanation:
Given data:
pH = 12.2
Hydroxide ion concentration = ?
Solution:
It is known that,
pH + pOH = 14
12.2 + pOH = 14
pOH = 14 - 12.2
pOH = 1.8
pOH = -log [OH⁻]
1.8 = -log [OH⁻]
10⁻¹°⁸ = [OH⁻]
[OH⁻] = 0.016
Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)
Given s = 7.4×10⁻³ M
So, Ksp is:
Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>
