The factors affect wave height is the duration of the wind
<span>The correct answer is Beryllium. They all belong to atoms that have low electronegativity, and as you go down the group the farther away you are the lower it gets. This means that they are not good at bonding with and attracting pairs of electrons. Berryllium has an electronegativity of 1.5, which is low compared to Fluorine, the highest one that has 4.</span>
Answer:
Volume of the solutions
This is the most important factor for her to control.
Titration involves acetic acid as the analyte and sodium hydroxide as the titrant. ph is more than 7 at the equivalency point.
The titration's equivalence point is the moment at which adequate titrant has been introduced to react to all of the material being titrated. A strong acid combined with something like a weak base produces a somewhat acidic salt. Similarly, combining a strong base with a mild acid yields a slightly basic salt.
When acetic acid combines with NaOH, it produces the salt sodium acetate and water. When acetic acid, CH3COOH, interacts with sodium hydroxide, sodium acetate, CH3COONa, and water are formed.
CH3COO- + H2O → CH3COOH + OH-
The acetate ion's reaction raises the ph to over 7.
Moles of acetic acid is equivalent to moles of sodium hydroxide at the equivalence point. As additional sodium hydroxide is added, the solution will become pink.
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Answer:
Option D, Concentration of NO2 decomposes after 4.00 s = 0.77 mol
Explanation:
Time (t) = 4.00\;s
Initial concentration of NO2 = 1.33 M
Integrated law for second order reaction:
![\frac{1}{[A]}=\frac{1}{[A]_0} =kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%3Dkt)
Where, [A] = Concentration after time, t
[A]0 = Intitial concentration, k = rate constant, t = time
On substituting values in the above
![\frac{1}{[A]}=\frac{1}{1.33} =0.255 \times 4.00](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D%5Cfrac%7B1%7D%7B1.33%7D%20%3D0.255%20%5Ctimes%204.00)
![\frac{1}{[A]} =1.772](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D1.772)
[A] = 0.5644 M
Concentration of NO2 decomposes after 4.00 s = 1.33 - 0.5644 = 0.7656 M
No. of mole = Molarity * volume
= 0.7656 * 1
= 0.7656 mol 0r 0.77 mol
