<h3><u>Answer;</u></h3>
<em>-49 °C</em>
<h3><u>Explanation and solution;</u></h3>
- Considering the fact that, the specific heat capacity of aluminum is 0.903 J/g x C, and the heat of vaporization of water at 25 C is 44.0 KJ/mol.
Moles water = 0.48 g / 18.02 g/mol
=0.0266 moles
<em>Heat lost by water</em> = 0.0266 mol x 44.0 kJ/mol
=1.17 kJ => 1170 J
<em>But heat lost =heat gained</em>
<em>Therefore;</em> Heat gained by aluminium = 1170 J
1170 = 55 x 0.903 ( T - 25) = 49.7 T - 1242
1170 + 1242 = 49.7 T
T = 48.5 °C ( 49 °C <em>at two significant figures)</em>
<em>Hence</em>, final temperature = 49 °C
Sodium , bromine zinc magnesium sulphur nitrogen potassium oxygen lead
The pKa of formic acid is 3.75. At pH of 5.00, b) [formate] > [formic acid].
Formic acid is a weak acid. Thus, together with its conjugate base (formate) they form a buffer system. We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.
![pH = pKa + log \frac{[formate]}{[formic\ acid]} \\\\5.00 = 3.75 + log \frac{[formate]}{[formic\ acid]}\\\\\frac{[formate]}{[formic\ acid]} = 17.8](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cfrac%7B%5Bformate%5D%7D%7B%5Bformic%5C%20acid%5D%7D%20%5C%5C%5C%5C5.00%20%3D%203.75%20%2B%20log%20%5Cfrac%7B%5Bformate%5D%7D%7B%5Bformic%5C%20acid%5D%7D%5C%5C%5C%5C%5Cfrac%7B%5Bformate%5D%7D%7B%5Bformic%5C%20acid%5D%7D%20%20%3D%2017.8)
As expected, at a pH above the pKa, the concentration of formate is higher than that of the formic acid.
The pKa of formic acid is 3.75. At pH of 5.00, b) [formate] > [formic acid].
Learn more: brainly.com/question/22821585