<u>Answer:</u> The pH of the buffer is 5.25
<u>Explanation:</u>
Let the volume of buffer solution be V
We know that:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjugate%20base%7D%5D%7D%7B%5Bacid%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.90
![[\text{conjugate base}]=\frac{2.25}{V}](https://tex.z-dn.net/?f=%5B%5Ctext%7Bconjugate%20base%7D%5D%3D%5Cfrac%7B2.25%7D%7BV%7D)
![[acid]=\frac{1.00}{V}](https://tex.z-dn.net/?f=%5Bacid%5D%3D%5Cfrac%7B1.00%7D%7BV%7D)
pH = ?
Putting values in above equation, we get:
Hence, the pH of the buffer is 5.25
The heat of the reaction, in kJ, when 4.18 g of the hydrocarbon are combusted 775.70 kJ.
The heat energy is given as :
q = m c ΔT + Ccal ΔT
q = ( 974 g× 4.184 ×6.9) + 624 ×6.9
q = 32424.59 J
moles of hydrocarbon = 0.0418 mol
heat of combustion = 32424.59 J / 0.0418 mol
= 775707.89 J
= 775.70 kJ
Thus, A 4.18 g sample of a hydrocarbon is combusted in a bomb calorimeter that contains 974 g of water. the temperature of the water increases by 6.9 °C when the hydrocarbon is combusted. the calorimeter constant for the calorimeter was determined to be 624 J/°C. what is the heat of the reaction is 775.70 kJ.
To learn more about calorimeter here
brainly.com/question/28943378
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There are four type of intermolecular forces: ionic, dipole-dipole, hydrogen bonds and London disperssion forces.
CH4 have no ions, so there are not ionic forces.
CH4 is a symetrical molecule, so there cannot be a net dipole in the molecule, so there is not dipole-dipole interaction.
Hydrogen bonding is only possbile when H is bonded to N, O or F, beacuse they are the atoms that considerable higher electgronegativy than hydrogen.
So, the only intermolecular force present in CH4 molecules is London disperssion forces, which is a force present in any molecule and is the weakiest one. That explains the low melting and boiling points of CH4.
Your answers would be k^+1, Cl2 and F-1
Hope this helps. Kinda winged it with my research.
