<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
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1 mol ------- 106g
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>
Answer:
3.711 L
Explanation:
The formula you need to use is the following:
3.4L / 298 K = V2 / 273 K
V2 = 3.711 L
Neutralization reaction is the reaction between acid and base.
There the neutralization reaction is:
HBr + LiOH ----------> LiBr + H2O.
Hope this helps!
The density of CO2 getting from experiment is 0.1/0.056 = 1.79 g/L. The percent error of this is (1.96 -1.79)/1.96*100%=8.67%. So the approximate percent error is 8.67%.
Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>
The process of preparing solutions from stock solutions of higher concentration is known as dilution.
Dilution is done with the aid of the dilution formula given below:
where
- C1 is the concentration of stock solution
- V1 is the volume of stock solution required to prepare a diluted solution
- C2 is the concentration of the diluted solution prepared
- V2 is the final volume of the diluted solution
From the data provided:
C1 is not given
V1 is unknown
C2 = 25%
V2 = 12 mL
- Assuming C1 is 50% solution
Volume of stock, V1, required is calculated as follows:
V1 = C2V2/C1
V1 = 25 × 12 /50
V1 = 6 mL
Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
Learn more about dilution formula at: brainly.com/question/7208546
