Answer:
Explanation:
<u>1) Reactants:</u>
The reactants are:
- <em>Molecular chlorine</em>: this is a gas diatomic molecule, i.e. Cl₂ (g)
- <em>Molecular fluorine</em>: this is also a gas diatomic molecule: F₂ (g)
<u>2) Stoichiometric coefficients:</u>
- <em>One volume of Cl₂ react with three volumes of F₂</em> means that the reaction is represented with coefficients 1 for Cl₂ and 3 for F₂. So, the reactant side of the chemical equation is:
Cl₂ (g) + 3F₂ (g) →
<u>3) Product:</u>
- It is said that the reaction yields <em>two volumes of a gaseous product;</em> then, a mass balance indicates that the two volumes must contain 2 parts of Cl and 6 parts of F. So, one volume must contain 1 part of Cl and 3 parts of F. That is easy to see in the complete chemical equation:
Cl₂ (g) + 3F₂ (g) → 2Cl F₃ (g)
As you see, that last equation si balanced: 2 atoms of Cl and 6 atoms of F on each side, and you conclude that the formula of the product is ClF₃.
Copper is the answer to ur question
Answer:
a. 0.182
b. 1.009
c. 1.819
Explanation:
Henderson-Hasselbach equation is:
pH = pKa + log [salt / acid]
Let's replace the formula by the given values.
a. 3 = 3.74 + log [salt / acid]
3 - 3.74 = log [salt / acid]
-0.74 = log [salt / acid]
10⁻⁰'⁷⁴ = 0.182
b. 3.744 = 3.74 + log [salt / acid]
3.744 - 3.74 = log [salt / acid]
0.004 = log [salt / acid]
10⁰'⁰⁰⁴ = 1.009
c. 4 = 3.74 + log [salt / acid]
4 - 3.74 = log [salt / acid]
0.26 = log [salt / acid]
10⁰'²⁶ = 1.819
I don't know which one I should answer
Answer:
Reagent A: PBr₃
Reagent B: Mg in Et₂O.
Explanation:
Hello,
In this case, your facing a problem in which a carboxylic acid is produced starting by an alcohol. More specifically, cyclopentanol must react with phosphorous tribromide in order to yield bromocyclopentane which is more likely to produce a carboxylic acid, therefore, reagent A is PBr₃.
On the other hand, by means of the production of the specified product, bromocyclopentane must react with carbon dioxide and magnesium in diethyl ether in acidic media to promote the production of the cyclopentanoic acid via the grignard reaction (substitution of the bromine by the carboxyle group), therefore, reagent B is Mg in Et₂O.
Best regards.
