Always remember that a compound can be separated into simpler substances by chemical methods/reactions. While elements cannot be broken down into simpler substances by chemical reactions. You can do a flame test and spectrum analysis to determine whether a solid material is an element or a compound. Check the boiling and/or melting point, color or density. Also check the solid material’s reaction with oxygen, hydrogen, calcium, or various acids. Examine and study its physical chemistry. The element(s) that may be present may be identified by checking the absorption edges from an x-ray spectrum.
NaN₃ is the chemical formula for Sodium Azide
Answer:
A.) 4.0
Explanation:
The general equilibrium expression looks like this:
![K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%20%7D)
In this expression,
-----> K = equilibrium constant
-----> uppercase letters = molarity
-----> lowercase letters = balanced equation coefficients
In this case, the molarity's do not need to be raised to any numbers because the coefficients in the balanced equation are all 1. You can find the constant by plugging the given molarities into the equation and simplifying.
<----- Equilibrium expression
![K = \frac{[2 M] [2 M]}{[1 M] [1 M] }](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5B2%20M%5D%20%5B2%20M%5D%7D%7B%5B1%20M%5D%20%5B1%20M%5D%20%7D)
<----- Insert molarities
<----- Multiply
<----- Divide
<u>Given information:</u>
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
<u>To determine:</u>
pH of 0.1 M NaF
<u>Explanation:</u>
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
