Answer:
NaCl: ionic, HF: hydrogen bond, HCl: dipole dipole , F2: dispersion force
Explanation:
complete question is:
The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion forces. Consider the compounds below, and classify each by its predominant attractive or intermolecular force among atoms or molecules of the same type.Identify each of the following ( NaCl, HF, HCl, F2) as Ionic, H Bonding, Dipole or Dispersion.
A simple chemical reaction with a single substrate shows a linear relationship between the rate of formation of product and the concentration of substrate, as shown below: ... The relationship between rate of reaction and concentration of substrate depends on the affinity of the enzyme for its substrate.
Answer:
52.45g
Explanation:
The computation of the mass of pure acetic acid in 125mL of this solution is shown below:
The percentage of mass would be equivalent to the g of solute in each 100g of water
As we know that
density = mass ÷ volume
So,
Volume = mass ÷ density
V = 100g / 1.049 (g / ml)
V = 95.328 mL
Now In every 95,328 ml of C_2H_4O_2 there are 40g of C_2H_4O_2
i.e.
each 125ml of C_2H_4O_2 there are 52.45g
SO,
x = 40g. 125ml ÷ 95.328
x = 52.45g
Answer:
Covalent Bond
Explanation:
In the diagram Carbon and each of the 4 Hydrogens are sharing electrons. They are also both non metals. Both of these are characteristics of Covalent Bonds.
Answer:
See explanation below
Explanation:
You are not providing the starting material, however, I manage to find a similar question to this, so I'm gonna use it as a basis to help you answer yours.
Now let's analyze what is happening in the reaction so we can predict the final product.
We have a ketone here, reacting at first with LDA. This is a very strong base that is commonly used in reactions with ketones and aldehydes to promove a condensation. To do this, as LDA is a strong base it will occur firts an acid base reaction, substracting the most acidic hydrogen in the molecule (Which in this case, is the Beta hydrogen of the carbonile). This will cause an enolate formation.
Then, this enolate will react with the CH3I and form a new product. The final result would be a ketone with a methyl group now attached. In the picture 2, you have the mechanism and final product.
Hope this helps