Answer:

Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
1. Assemble all the data in one place, with molar masses above the formulas and other information below them.
Mᵣ: 58.44
NaCl + AgNO₃ ⟶ NaNO₃ + AgCl
m/g: 0.245
V/mL: 50.
c/mmol·mL⁻¹: 0.0180
2. Calculate the moles of each reactant

3. Identify the limiting reactant
Calculate the moles of AgCl we can obtain from each reactant.
From NaCl:
The molar ratio of NaCl to AgCl is 1:1.

From AgNO₃:
The molar ratio of AgNO₃ to AgCl is 1:1.

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.
4. Calculate the moles of excess reactant
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
I/mmol: 0.900 4.192 0
C/mmol: -0.900 -0.900 +0.900
E/mmol: 0 3.292 0.900
So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.
5. Calculate the concentration of Cl⁻
![\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BCl%24%5E%7B-%7D%24%5D%20%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B3.292%20mmol%7D%7D%7B%5Ctext%7B50.%20mL%7D%7D%20%3D%20%5Ctextbf%7B0.066%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20concentration%20of%20chloride%20ion%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.066%20mol%2FL%7D%7D%24%7D)
Answer: The average atomic mass of oxygen is 15.999 amu
Explanation:
Mass of isotope O-16 = 15.995 amu
% abundance of isotope O-16= 99.759 % = 
Mass of isotope O-17 = 16.995 amu
% abundance of isotope O-17 = 0.037% = 
Mass of isotope O-18 = 17.999 amu
% abundance of isotope O-18 = 0.204% = 
Formula used for average atomic mass of an element :

![A=\sum[(15.995\times 0.99759)+(16.995\times 0.00037)+(17.999 \times 0.00204)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2815.995%5Ctimes%200.99759%29%2B%2816.995%5Ctimes%200.00037%29%2B%2817.999%20%5Ctimes%200.00204%29%5D)

Thus the average atomic mass of oxygen is 15.999 amu
I think it’s b because in the question it says that the agro technician ran some tests on the soil and found out that it’s acidic and then after that the farmer should discuss how to remove the acid so the soil can be fit for farming
heh.......... sorry man............... but this was posted over 5 hours ago........ so nobody is gonna see it and you probably dont need the answer anymore..... so errrr..... imma justtttt..... take these points :D