Answer:
8.2×10¯⁵ L
Explanation:
82 μL.
We can convert 82 μL to L by doing the following:
To convert from microlitre (μL) to litre(L), we need to know how many microlitre (μL) that makes up a litre (L)
Recall:
1 μL = 1×10¯⁶ L
Therefore,
82 μL = 82 μL × 1×10¯⁶ L/1 μL
82 μL = 8.2×10¯⁵ L
Therefore, 82 μL is equivalent to 8.2×10¯⁵ L
Answer:
Yes, it is due to the gravitational pull from the moon
Explanation:
The mass of Garfield didn't change during the flight, which is to say, the amount of matter in his body is the same on Earth as on the Moon.
However, <em>weight is a measurement of how strongly mass is pulled towards the ground, due to the gravitational pull.</em>
So even though Garfield's mass is the same, the weight is less due to the lower gravitational pull from the Moon compared to Earth (the weight is one-sixth of the original weight as gravity on the moon is about one-sixth of Earth's). When Garfield returns to Earth, the scale will still read 24 lbs.
The answer is going to be b. hope that helped
Answer: [H3O+] > [OH-] and Kw = 1 x 10 -14
Explanation:
i just had this question and tried to look it up and i got it right by guessing my bad but i hope this is ur answer as well
Answer:
The pH of the solution is 4.282
Explanation:
First we need to write the dissociation reaction
⇄ ![CLO^-_{(aq)} +H_3O^+_{(aq)}](https://tex.z-dn.net/?f=CLO%5E-_%7B%28aq%29%7D%20%2BH_3O%5E%2B_%7B%28aq%29%7D)
Also, we need to prepare I C E table
⇄ ![CLO^-_{(aq)} +H_3O^+_{(aq)}](https://tex.z-dn.net/?f=CLO%5E-_%7B%28aq%29%7D%20%2BH_3O%5E%2B_%7B%28aq%29%7D)
I 0.078 0 0
C - x +x +x
E 0.078 - x x x
Thus, pH= - Log[H₃O⁺]
The equilibrium concentration of H₃O⁺ = x
So, we solve for x
![K_a = \frac{[CLO^-].[H_3O^+]}{[HCLO]} \\\\3.5X10^{-8} = \frac{[x].[x]}{[0.078-x]} \\\\3.5X10^{-8} =\frac{x^2}{0.078-x} \\\\x^2 = 3.5X10^{-8} (0.078-x)\\\\x^2 = 2.73X10^{-9} - (3.5X10^{-8})x\\\\x^2 + (3.5X10^{-8})x - 2.73X10^{-9} =0](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BCLO%5E-%5D.%5BH_3O%5E%2B%5D%7D%7B%5BHCLO%5D%7D%20%5C%5C%5C%5C3.5X10%5E%7B-8%7D%20%3D%20%5Cfrac%7B%5Bx%5D.%5Bx%5D%7D%7B%5B0.078-x%5D%7D%20%5C%5C%5C%5C3.5X10%5E%7B-8%7D%20%3D%5Cfrac%7Bx%5E2%7D%7B0.078-x%7D%20%5C%5C%5C%5Cx%5E2%20%3D%203.5X10%5E%7B-8%7D%20%280.078-x%29%5C%5C%5C%5Cx%5E2%20%3D%202.73X10%5E%7B-9%7D%20-%20%283.5X10%5E%7B-8%7D%29x%5C%5C%5C%5Cx%5E2%20%2B%20%283.5X10%5E%7B-8%7D%29x%20-%202.73X10%5E%7B-9%7D%20%3D0)
Solving this quadratic equation; x = 5.2232 x 10⁻⁵ M
pH = - Log[H₃O⁺]
pH = - Log[5.2232 x 10⁻⁵]
pH = 4.282 (in 3 decimal places)
Therefore, the pH of the solution is 4.282