The answer is Non-renewable
Answer:
![[Cl^-]=232.3\frac{mgCl^-}{L}](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D232.3%5Cfrac%7BmgCl%5E-%7D%7BL%7D)
Explanation:
Hello,
In this case, we can represent the chemical reaction as:

In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:

Finally, for the given volume of water in liters (0.100L), we compute the required concentration:
![[Cl^-]=\frac{23.2mgCl^-}{0.100L}\\](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D%5Cfrac%7B23.2mgCl%5E-%7D%7B0.100L%7D%5C%5C)
![[Cl^-]=232.3\frac{mgCl^-}{L}](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D232.3%5Cfrac%7BmgCl%5E-%7D%7BL%7D)
Best regards.
Answer:
109.34 g
Explanation:
2NaOH(aq) + H2SO4(aq) ------> Na2SO4(aq) + 2H2O(l)
Number of moles of NaOH = 105g/40g/mol = 2.6 moles
From the reaction equation;
2 moles of NaOH yields 1 mole of sodium sulphate
2.6 moles of NaOH yields = 2.6 × 1/2 = 1.3 moles of sodium sulphate
Number of moles of H2SO4= 75.5g/98 g/mol = 0.77 moles
From the reaction equation;
1 mole of H2SO4 yields 1 mole of sodium sulphate
Hence, 0.77 moles of H2SO4 yields 0.77 moles of sodium sulphate
So H2SO4 is the limiting reactant.
Theoretical yield = number of moles × molar mass
= 0.77 mol ×142 g/mol
= 109.34 g
Answer:
the answer is d. 74.02
Explanation:
When ogyen all together with the different things to get this.