Answer:
49.4 g Solution
Explanation:
There is some info missing. I think this is the original question.
<em>A chemistry student needs 20.0g of acetic acid for an experiment. He has 400.g available of a 40.5 % w/w solution of acetic acid in acetone. </em>
<em>
Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.</em>
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We have 400 g of solution and there are 40.5 g of solute (acetic acid) per 100 grams of solution. We can use this info to find the mass of acetic acid in the solution.
Since we only need 20.0 g of acetic acid, there is enough of it in the solution. The mass of solution that contains 20.0 g of solute is:
Answer:
Option C :
a chemical formula that shows the relative number of each type of atom in a molecule, using the smallest possible ratio
Explanation:
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
Tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule
For Example
C₂H₆O₂ Consist of Carbon (C), Hydrogen (H), and Oxygen (O)
Now
Look at the ratio of these three atoms in the compound
C : H : O
2 : 6 : 2
Divide the ratio by two to get simplest ratio
C : H : O
2/2 : 6/2 : 2/2
1 : 3 : 1
So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1
So the empirical formula will be
Empirical formula of C₂H₆O₂ = CH₃O
So, Option C is correct :
a chemical formula that shows the relative number of each type of atom in a molecule, using the smallest possible ratio
You multiply the number of atoms by 12 to get how many electrons (since each atom has 12 electrons in it)
you multiply the number of atoms by 13 to get how many neutrons
(since each atom of this isotope has 13 neutrons in it)
Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A
