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Svetach [21]
3 years ago
7

What would be the total volume of the new solution when it is changed from 0.2 M to 0.04 M?

Chemistry
2 answers:
34kurt3 years ago
6 0
The question is incomplete.

You need two additional data:

1) the original volume
2) what solution you added to change the volume.

This is a molarity problem, so remember molarity definition and formula:

M = n / V in liters: number of moles per liter of solution

To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).

The original solution was:

V= 1 ml
M = 0.2 M

Using the formula for molarity, M = n / V

n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles

For the final solution:

n = 0.0002 moles
M = 0.04

From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l

Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml.  This would be the answer for the hypothetical problem that I assumed for you.

I hope this gives you all the cues you need to answer similar problems about molarity.
konstantin123 [22]3 years ago
5 0

Answer:

The answer is 50 mL

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quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3
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The limiting reactant is acetic acid. All 125 g will react.  

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

Mᵣ:                 60.05            78.00

                3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g:           125               275  

2. Calculate the moles of each reactant

\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}

3. Calculate the moles of (CH₃COO)₃Al from each reactant

\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} =  \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol  (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol  (CH$_{3}$COO)$_{3}$Al}

\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}

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