Answer:
D = 4 g/mL
General Formulas and Concepts:
Density = Mass over Volume
Explanation:
<u>Step 1: Define</u>
m = 100 g
V = 25 mL
D = unknown
<u>Step 2: Find D</u>
- Substitute: D = 100 g/25 mL
- Evaluate: D = 4 g/mL
Answer:
C) METALLIC IS THE CORRECT ANSWER
Explanation: I just did the exam
Answer:
double replacement
synthesis
double replacement
i think this one is decomposition
synthesis
double replacement
single replacement
single replacement
double replacement
single replacement .....
hopefully i help
Answer:
pH change is -0.07
Explanation:
Using H-H equation for acetic acid:
pH = pKa + log [Acetate salt] / [Acetic acid]
Replacing:
pH = 4.74 + log[1.188M] / [1.188M]
pH = 4.74
The HCl reacts with sodium acetate producing acetic acid, thus:
HCl + CH₃COONa → CH₃COOH + NaCl
That means the final moles of sodium acetate are initial moles - moles of HCl and moles of acetic acid are initial moles + moles of HCl.
As the volume of the buffer is 1.0L, initial moles of both substances are 1.188moles. After reaction, the moles are:
sodium acetate: 1.188mol - 0.1mol = 1.088mol
Acetic acid: 1.188mol + 0.1mol = 1.288mol
Using again H-H equation:
pH = 4.74 + log[1.088M] / [1.288M]
pH = 4.67
pH change is: 4.67 - 4.74 = -0.07
Conjugate base of Propanoic acid (
is propanoate where -COOH group gets converted to -CO
. The structure of conjugate base of Propanoic acid is shown in the diagram.
The 
above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.
= 
+ log![\frac{[Conjugate base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BConjugate%20base%5D%7D%7B%5Bacid%5D%7D)
=4.9+log
=5.85
As 90% conjugate base is present, so propanoic acid present 10%.
