Answer:
A substance that is composed only of atoms having the same atomic number is ... 36 grams of an unknown liquid at its boiling point,.
Answer:
1.) 0.1 M
2.) 0.2 M
3.) 1 M
4.) Solution #3 is the most concentrated because it has the highest molarity. This solution has the largest solute to solvent ratio. The more solvent there is, the lower the concentration and molarity.
Explanation:
To find the molarity, you need to (1) convert grams NaOH to moles (via molar mass from periodic table) and then (2) calculate the molarity (via the molarity equation). All of the answers should have 1 sig fig to match the given values.
Molar Mass (NaOH): 22.99 g/mol + 16.00 g/mol + 1.008 g/mol
Molar Mass (NaOH): 39.998 g/mol
4 grams NaOH 1 mole
---------------------- x ------------------ = 0.1 moles NaOH
39.998 g
1.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (1 L)
Molarity = 0.1 M
2.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (0.5 L)
Molarity = 0.2 M
3.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (0.1 L)
Molarity = 1 M
Answer:
Argon has 8 valence electrons and no extras, it does not require a bond in order to fill its shells, its satisfied by itself.
Chlorine is missing 1 Electron, if it connects with another Chlorine it will be satisfying both of their needs with a Covalent bond.
Explanation:
I believe it is 65.37.
Let me know if this is correct. Also good luck!!
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)
