Answer: The pressure of the tires after driving is 4.9atm
Explanation: Please see the attachments below
The solubility of glucose at 30°C is 125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water a
2 answers:
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Answer : The pressure in the flask after reaction complete is, 2.4 atm
Explanation :
To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.
where,
P = final pressure in the flask = ?
R = gas constant = 0.0821 L.atm/mol.K
T = temperature =
V = volume = 4.0 L
= moles of
= 0.20 mol
= moles of
= 0.20 mol
Now put all the given values in the above expression, we get:
Thus, the pressure in the flask after reaction complete is, 2.4 atm
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!