Answer is: pH of hydroxylamine solution is 9,23.
Kb(NH₂OH) = 1,8·10⁻⁵<span>.
c</span>₀(NH₂OH)<span> = 0,0500 M =
0,05 mol/L.
c(NH</span>₂⁺) = c(OH⁻) = x.
c(NH₂OH<span>) = 0,05 mol/L - x.
Kb = c(NH</span>₂⁺) · c(OH⁻) / c(NH₂OH).
0,0000000066 = x² / (0,05 mol/L - x).
solve quadratic equation: x = c(OH⁻) = 0,000018 mol/L.<span>
pOH = -log(</span>0,000018 mol/L) = 4,74.<span>
pH = 14 - 4,74 = 9,23.</span>
In a food chain we arrange the energy in the form of a pyramid.
The producers are on the base of pyramid and then consumers are towards peak.
in the given food chain grass is being eaten by grasshopper which are food of birds.
Grasshoppers are also eaten up by Hawks. so both brids and hawks are feeding upon grasshoppers thus the amount of energy transferred from the grass to the grasshopper is the same as the amount of energy transferred from the grasshopper to the bird.
Answer:
Helium is the second element on the periodic table. It is located in period 1 and group 18 or 8A on the righthand side of the table. This group contains the noble gases, which are the most chemically inert elements on the periodic table. Each He atom has two protons and usually two neutrons and two electrons.
Explanation:
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Answer : All of the above are valid expressions of the reaction rate.
Explanation :
The given rate of reaction is,
The expression for rate of reaction for the reactant :
![\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DNH_3%3D-%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DO_2%3D-%5Cfrac%7B1%7D%7B7%7D%5Ctimes%20%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The expression for rate of reaction for the product :
![\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DNO_2%3D%2B%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DH_2O%3D%2B%5Cfrac%7B1%7D%7B6%7D%5Ctimes%20%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D)
From this we conclude that, all the options are correct.
