If it takes three "breathes" to blow up a balloon to 1.2 L, and each breath supplies the balloon with 0.060 moles of exhaled air
1 answer:
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Answer:
a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂
b) Ni(OH)₂
c) KOH
d) 0.927 g
e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M
Explanation:
a) The equation is:
2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)
b) The precipitate formed is Ni(OH)₂
c) The limiting reactant is:
From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:
Hence, the limiting reactant is KOH.
d) The mass of the precipitate formed is:
e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:
I hope it helps you!
Hello!
To solve this problem we are going to use the Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is 4,76:
![pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%20%29%20)
![\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_3COOH%5D%7D%7B%5BCH_3COONa%7D%3D%2010%5E%7B%28pH-pKa%29%5E%7B-1%7D%7D%3D10%5E%7B%284-4%2C76%29%5E%7B-1%7D%7D%3D5%2C75%20)
So, the mole ratio of CH₃COOH to CH₃COONa is 5,75
Have a nice day!
To solve this problem we are going to use the Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is 4,76:
So, the mole ratio of CH₃COOH to CH₃COONa is 5,75
Have a nice day!