Explanation:
It is known that 
value of acetic acid is 4.74. And, relation between pH and is as follows.
pH = pK_{a} + log ![\frac{[CH_{3}COOH]}{[CH_{3}COONa]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOH%5D%7D%7B%5BCH_%7B3%7DCOONa%5D%7D)
= 4.74 + log 
So, number of moles of NaOH = Volume × Molarity
= 71.0 ml × 0.760 M
= 0.05396 mol
Also, moles of 
= moles of 
= Molarity × Volume
= 1.00 M × 1.00 L
= 1.00 mol
Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.
Initial : 1.00 mol 1.00 mol
NaoH addition: 0.05396 mol
Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)
= 0.94604 mol = 1.05396 mol
As, pH = pK_{a} + log ![\frac{[CH_{3}COONa]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOONa%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
= 4.74 + log 
= 4.69
Therefore, change in pH will be calculated as follows.
pH = 4.74 - 4.69
= 0.05
Thus, we can conclude that change in pH of the given solution is 0.05.
Answer:
2.765amu is the contribution of the X-19 isotope to the weighted average
Explanation:
The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:
X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21
The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:
19.00amu * 0.1455 =
2.765amu is the contribution of the X-19 isotope to the weighted average
Well depends how fast they're going if it's a slow speed a bus but at a fast speed a bike because you wanna be careful while stopping
The chemist the count the number of particles (Atoms, Molecules or Formula Unit) in a given number of moles of a substance by using following relationship.
Moles = # of Particles / 6.022 × 10²³
Or,
# of Particles = Moles × 6.022 × 10²³
So, from above relation it is found that 1 mole of any substance contains exactly 6.022 × 10²³ particles. Greater the number of moles greater will be the number of particles.
