I believe the answer is C
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
Explanation:
Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²
2x 0.007 50 + x
![K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BAg%24%5E%7B%2B%7D%24%5D%24%5E%7B2%7D%24%5BCO%24_%7B3%7D%5E%7B2-%7D%24%5D%7D%20%3D%20%282x%29%5E%7B2%7D%5Ctimes%200.00750%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5C0.0300x%5E%7B2%7D%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5Cx%5E%7B2%7D%20%3D%202.70%20%5Ctimes%2010%5E%7B-10%7D%5C%5Cx%20%3D%20%5Csqrt%7B2.70%20%5Ctimes%2010%5E%7B-10%7D%7D%20%3D%20%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20maximum%20concentration%20of%20Ag%24%5E%7B%2B%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B-5%7D%7D%5Ctextbf%7B%20mol%2FL%20%7D%7D%24%7D)
Answer:
Your body temperature would decrease
Explanation:
