6. What is the molarity of 1.35 mol H2So4 in 245 mL of solution?
1 answer:
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Answer:
particles of gold
Explanation:
To convert the number of moles of any substance, in this case gold, you need Avogadro's number.
Avogadro's number is always ×
moles Au ×
=
particles of gold
Hello!
To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:
![pH=pKa+log( \frac{[A^{-}] }{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%20%7D%7B%5BHA%5D%7D%20%29)
Now, we use the definition of pH and clear [H₃O⁺] from there:
![pH=-log[H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M](https://tex.z-dn.net/?f=%20%5BH_3O%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-pH%7D%20%3D10%5E%7B-3%2C84%7D%3D0%2C00014%20M)
So, the [H₃O⁺] concentration is 0,00014 M
Have a nice day!
To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:
Now, we use the definition of pH and clear [H₃O⁺] from there:
So, the [H₃O⁺] concentration is 0,00014 M
Have a nice day!