The molarity of the solution of H₃PO₄ needed to neutralize the KOH solution is 0.35 M
<h3>Balanced equation </h3>
H₃PO₄ + 3KOH —> K₃PO₄ + 3H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 1
- The mole ratio of the base, KOH (nB) = 3
<h3>How to determine the molarity of H₃PO₄ </h3>
- Volume of acid, H₃PO₄ (Va) = 10.2 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.2 M
- Volume of base, Ca(OH)₂ (Vb) = 53.5 mL
- Molarity of acid, H₃PO₄ (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 10.2) / (0.2 × 53.5) = 1 / 3
(Ma × 10.2) / 10.7 = 1 / 3
Cross multiply
Ma × 10.2 × 3 = 10.7
Ma × 30.6 = 10.7
Divide both side by 30.6
Ma = 10.7 / 30.6
Ma = 0.35 M
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Decreasing the turns of wire
Complete Question:
check the first image for complete part of the question
Answer and Explanation:
Epoxide is a three membered ring made up of two carbon atoms and one oxygen atom. Epoxides are cyclic ethers. Due to its ring size, it is highly strained and very reactive. Epoxide ring opening takes place with respect to addition of acid and base.
Ring opening of epoxide with acid:
In the presence of base, the nucleophile attacks the epoxide ring at more substituted site and inverse stereochemistry takes place.(check file 2 attached)
Ring opening of epoxide with base:
The backside attack of nucleophile takes place in less substituted site and then it undergoes protonation to form a product.
(check file 2 attached)
Answer:
7 hours
Explanation: step 1. 160x4=640
step 2. 1180-640=540
step 3. 540÷180=3
step 4. 3+4=7
Answer:
Just show them their place.
Explanation:
hope this helps