) The only noble gas without eight valence electrons is __________. A) Ar B) Ne C) Kr D) He E) All noble gases have eight valenc
1 answer:
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<u>Answer:</u> The specific heat of metal is 2.34 J/g°C
<u>Explanation:</u>
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 35 mL
Putting values in above equation, we get:
When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.
The equation used to calculate heat released or absorbed follows:
......(1)
where,
q = heat absorbed or released
= mass of metal = 4.82 g
= mass of water = 35 g
= final temperature = 34.5°C
= initial temperature of metal = 115°C
= initial temperature of water = 28.7°C
= specific heat of metal = ?
= specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
Hence, the specific heat of metal is 2.34 J/g°C
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!