The beaker of acetic acid will cool more quickly.
The specific heat capacity of acetic acid is about half that of water.
Thus, it takes twice as much heat gain (or loss) in acetic acid to cause a given change in temperature.
If everything else is constant and heat is being lost at the same rate, the temperature of the acetic acid should drop twice as fast as that of water.
Answer:
4.13X10^3= 4130 in the expanded form
The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.
<h3>What is the value of Van t Hoff factor?</h3>
For most non-electrolytes dissolved in water, the Van 't Hoff factor is essentially $ 1 $ . For most ionic compounds dissolved in water, the Van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
<h3>Which has highest Van t Hoff factor?</h3>
The Van't Hoff factor will be highest for
A. Sodium chloride.
B. Magnesium chloride.
C. Sodium phosphate.
D. Urea.
Learn more about van't off factor here:
<h3>
brainly.com/question/22047232</h3><h3 /><h3>#SPJ4</h3>
It’s charge was neutral due to the equal number of protons and electrons. when it becomes an ion it loses 3 electrons leaving behind only 10. the answer is 10. the equation is +13 +(-10)=+3
Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
<span>Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; </span>
<span>At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; </span>
<span>so 0.17 = x² / (0.20 - x); </span>
<span>Solving for x using the quadratic formula: </span>
<span>x = [H+] = 0.063 M or pH = - log [H+] = 1.2.</span>
