Answer:
pH = 4.8
Explanation:
A buffer is formed by a weak acid (0.145 M HC₂H₃O₂) and its conjugate base (0.202 M C₂H₃O₂⁻ coming from 0.202 M KC₂H₃O₂). The pH of a buffer system can be calculated using Henderson-Hasselbalch's equation.
![pH = pKa + log\frac{[base]}{[acid]} \\pH = -log(1.8 \times 10^{-5} )+log(\frac{0.202M}{0.145M} )\\pH=4.8](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%5C%5CpH%20%3D%20-log%281.8%20%5Ctimes%2010%5E%7B-5%7D%20%29%2Blog%28%5Cfrac%7B0.202M%7D%7B0.145M%7D%20%29%5C%5CpH%3D4.8)
Answer:
Hydrogen, H_2
Explanation:
mass of each gas is 10.0 g
number of mole = mass/ molar mass
number of moles is directly proportional to volume at constant temp and pressure
this implies that the volume is inversely proportional to molar mass. And Among all the gases in periodic table the molar mass of Hydrogen is the least.
molar mass of H2=2 g/mol
Since, H2 has minimum molar mass then for the same mass of the gases Hydrogen will have maximum volume.
Answer:
C. Chemical change
Explanation:
A physical change is where something is changed but it doesnt affect the build up of the chemical. For example, if you broke sticks and threw them on the ground, that would be a physical change because the change is happening to the physical being of the object and not its chemical buildup. However, if you lit those sticks on fire, that would be considered a chemical change because you end up with two substances, ash and the remnants of the stick. A nuclear reaction would result in something blowing up so its not that. And a physical property is like what it looks like or how it smells. Hope I helped you!
The decomposition time : 7.69 min ≈ 7.7 min
<h3>Further explanation</h3>
Given
rate constant : 0.029/min
a concentration of 0.050 mol L to a concentration of 0.040 mol L
Required
the decomposition time
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time
For first-order reaction :
[A]=[Ao]e^(-kt)
or
ln[A]=-kt+ln(A0)
Input the value :
ln(0.040)=-(0.029)t+ln(0.050)
-3.219 = -0.029t -2.996
-0.223 =-0.029t
t=7.69 minutes
