The correct answer is a Photon.
One photon is released for each event. Photons are elementary particles of all electromagnetic radiation, including light.
First, we need to get moles of NaOH:
when moles NaOH = volume * molarity
= 0.02573L * 0.11 M
= 0.0028 moles
from the reaction equation:
H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)
we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH
∴ X mol H3PO4 reacts with → 0.0028 moles NaOH
∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol
now we can get the concentration of H3PO4:
∴[H3PO4] = moles H2PO4 / volume
= 9.4 x 10^-4 / 0.034 L
= 0.028 M
Answer:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
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Answer:
Calculate the molar concentration of the NaOH solution that you prepared Number of moles of KHP = Number of moles NaOH = 2.476 x 10 -3 moles Number of moles NaOH = Mb x Vb Mb = 2.476 x 10 -3 moles / 0.0250 L (equivalence point) = 0.0990 M 3
Explanation:
