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:
0.0187 M
Explanation:
Step 1: Write the balanced neutralization reaction
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of HCl
18.7 mL of 0.01500 M HCl react.
0.0187 L × 0.01500 mol/L = 2.81 × 10⁻⁴ mol
Step 3: Calculate the reacting moles of NaOH
The molar ratio of HCl to NaOH is 1:1. The reacting moles of NaOH are 1/1 × 2.81 × 10⁻⁴ mol = 2.81 × 10⁻⁴ mol.
Step 4: Calculate the molarity of NaOH
2.81 × 10⁻⁴ moles are in 15.00 mL of NaOH.
[NaOH] = 2.81 × 10⁻⁴ mol/0.01500 L = 0.0187 M
The Cascades rain shadow can be described as such: ocean-influenced moist air masses are forced to rise when they meet the tall moun- tains. The rising air cools, condenses, and the moisture falls as precipitation. On the leeward (dry) side of the mountain, the now dry air warms and sinks.
An ionic compound is formed between a metal and a non metal. The metal being electropositive can form a cation by transferring the electron to the electronegative non metal that gains the electrons to form an anion. Both the elements try to get a stable octet configuration by the transfer of electrons. The number of electrons lost by metal will be equal to the number of electrons gained by the non metal. Hence, the magnitude of positive charge on the cation will be equal to the magnitude of negative charge on the anion. Therefore, the overall charge on the compound will be 0
So the correct answer is the sum of all charges in the formula for an ionic compound is 0
Due to increase the surface area of a solute.
Hope this helps have a very good dayyyyyy.........
