Question

what is the final concentration of a solution prepared by diluting 25.0 ml of 12.0 m hcl to a final volume of 1.20 l?
0.250 m
250 m
0.00250 m
0.360 m
question 29 (0.0862 points)
what is the name of the functional group in this molecule?
h-c=o
/
o-h
ether
ketone
alcohol
carboxylic acid

Explanation:

First Question (Solution Concentration)

Step1: Recall Dilution Formula

The dilution formula is \( C_1V_1 = C_2V_2 \), where \( C_1 \) is initial concentration, \( V_1 \) initial volume, \( C_2 \) final concentration, \( V_2 \) final volume.

Step2: Convert Units

\( V_1 = 25.0 \, \text{mL} = 0.0250 \, \text{L} \), \( C_1 = 12.0 \, \text{M} \), \( V_2 = 1.20 \, \text{L} \).

Step3: Solve for \( C_2 \)

Rearrange formula: \( C_2 = \frac{C_1V_1}{V_2} \). Substitute values: \( C_2 = \frac{12.0 \, \text{M} \times 0.0250 \, \text{L}}{1.20 \, \text{L}} \). Calculate: \( \frac{0.300}{1.20} = 0.250 \, \text{M} \).

The molecule has \( -\text{COOH} \) (carboxylic acid) structure (carbonyl \( \text{C}=\text{O} \) and hydroxyl \( -\text{OH} \) on same carbon). Ether has \( -\text{O}- \), ketone has \( \text{C}=\text{O} \) between two carbons, alcohol has \( -\text{OH} \) (not with carbonyl). So it's carboxylic acid.

Answer:

0.250 M

Second Question (Functional Group)