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)
Complete question is;
A drop of water has a volume of approximately 7 × 10⁻² ml. How many water molecules does it contain? The density of water is 1.0 g/cm³.
This question will require us to first find the number of moles and then use avogadro's number to get the number of water molecules.
<em><u>Number of water molecules = 2.34 × 10²¹ molecules</u></em>
We are given;
Volume of water; V = 7 × 10⁻² ml
Density of water; ρ = 1 g/cm³ = 1 g/ml
Formula for mass is; m = ρV
m = 1 × 7 × 10⁻²
m = 7 × 10⁻² g
from online calculation, molar mass of water = 18.01 g/mol
Number of moles(n) = mass/molar mass
Thus;
n = (7 × 10⁻²)/18.01
n = 3.887 × 10⁻³ mol
from avogadro's number, we know that;
1 mol = 6.022 × 10²³ molecules
Thus,3.887 × 10⁻³ mol will give; 6.022 × 10²³ × 3.887 × 10⁻³ = 2.34 × 10²¹ molecules
Read more at; brainly.in/question/17990661
An intensive property is a property that does not change depending on how much mass of it you are considered. An example of an intensive property is density. No matter how much water you examine, the density of the sample will be 1g/cm³.
It's most natural state has a charge of -2. So, a negative ion with two more electrons than is normal.
Answer:
1) 1.15 mol
2) M=0.45
3) 22.5 mL
4) 6.25 mL
Explanation:
1)
550 mL= 0.55 L
M= mol solute/ L solution
mol solute= M * L solution
mol solute= (2.1 M * 0.55 L ) M=1.15 mol solute
2)
155 mL = 0.155 L
80 g -> 1 mol NH4NO3
5.61 g -> x
x= (5.61 g * 1 mol NH4NO3)/80 g x= 0.07 mol NH4NO3
M=(0.07 mol NH4NO3)/0.155 L M=0.45
3) M1V1=M2V2
V1= M2V2/M1
V1= (0.500 M * 0.225 L)/5.00 M V1=0.0225 L =22.5 mL
4) M1V1=M2V2
V1= M2V2/M1
V1= (0.25 M * 0.45 L)/ 18.0 M
V1=6.25 x 10^-3 L = 6.25 mL
