Answer:
2 molecules of water represents 3.32 x 10^-24 moles of water.
Explanation:
To find the solution to this problem, you have to use the concept of Avogadro´s number, that is in 1 mol of any element o compound there are 6.022 x 10^23 molecules. Then,
1 mol H2O ------------- 6.022 x 10^23 molecules
x= 3.32 x 10^-24 ---- 2 molecules.
2 molecules of water represents 3.32 x 10^-24 moles of water.
Given :
Volume of NaCl solution 2.5 L .
Molarity of NaCl solution is 0.070 M .
To Find :
How many moles are present in the solution.
Solution :
Let, n be the number of moles.
We know, molarity is given by :

So,

Therefore, number of moles of NaCl is 0.175 moles.
Answer:
The answer to your question is 88.7 ml
Explanation:
Data
Volume = ?
Concentration of NaOH = 0.142 M
Volume of H₂C₄H₄O₆ = 21.4 ml
Concentration of H₂C₄H₄O₆ = 0.294 M
Balanced chemical reaction
2 NaOH + H₂C₄H₄O₆ ⇒ Na₂C₄H₄O₆ + 2H₂O
1.- Calculate the moles of H₂C₄H₄O₆
Molarity = moles/volume
Solve for moles
moles = Molarity x volume
Substitution
moles = 0.294 x 21.4/1000
Result
moles = 0.0063
2.- Use proportions to calculate the moles of NaOH
2 moles of NaOH ------------------ 1 moles of H₂C₄H₄O₆
x ------------------ 0.0063 moles
x = (0.0063 x 2) / 1
x = 0.0126 moles of NaOH
3.- Calculate the volume of NaOH
Molarity = moles / volume
Solve for volume
Volume = moles/Molarity
Substitution
Volume = 0.0126/0.142
Result
Volume = 0.088 L or 88.7 ml
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)