Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol.
<span>[mW (m) - mW (g)]/mW (m) x 100% </span>
<span>(they want % error so, if it is negative, just get rid of the sign) </span>
The pH = 2.41
<h3>Further explanation</h3>
Given
5.0% by mass solution of acetic acid
the density of white vinegar is 1.007 g/cm3
Required
pH
Solution
Molarity of solution :

Ka for acetic acid = 1.8 x 10⁻⁵
[H⁺] for weak acid :
![\tt [H^+]=\sqrt{Ka.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D)
Input the value :
![\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.839%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D0.00388%3D3.88%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~3.88%3D2.41)
These 5 metals are: zinc,copper,tin,bronze,nickel
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano


1 gramo de metano aporta 50.125 kilojoules.
Octano


1 gramo de metano aporta 48.246 kilojoules.