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3 years ago

Calculate the volume of a 2.30 mol/L solution of Na2CO3 that contains 0.325 mol of solute.

Chemistry

1 answer:

Blababa [14]3 years ago

5 0

Answer:

141 mL is the volume of a 2.30 mol/L solution of Na₂CO₃ that contains 0.325 mol of solute.

Explanation:

Molarity is a measure of concentration of a solution that indicates the amount of moles of solute that appear dissolved in each liter of the mixture.

So 2.30 mol / L indicates that 2.30 moles dissolve in 1 L of solution. You want to know in what volume is contained 0.325 mol of solute. For this you apply the following rule of three: if 2.30 moles of solute are present in 1 L of solution, 0.325 moles in how much volume is it?

$volume=\frac{0.325 moles*1L}{2.30 moles}$

volume= 0.141 L

Being 1L=1000 mL, then 0.141 L=141 mL

<u><em>141 mL is the volume of a 2.30 mol/L solution of Na₂CO₃ that contains 0.325 mol of solute.</em></u>

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Most reactions, including enzyme-catalyzed reactions, proceed faster at higher temperatures. However, for a given enzyme, the ra

Tju [1.3M]

Answer:

It has denatured

Explanation:

When the temperature get high the enzymes tend to change shape and denaturing occurs.

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3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

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3 years ago

Calculate the mass of iron oxide that contains a trillion iron atoms

4vir4ik [10]

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3 years ago

Why do group 1 cations form precipitates when mixed with hcl?

Xelga [282]

Both of you are overlooking a pretty big component of the question...the Group I cation isn't being dissociated into water. We're testing the solubility of the cation when mixed with HCl. And this IS a legitimate question, seeing as our lab manual is the one asking.

<span>By the way, the answer you're looking for is "Because Group I cations have insoluble chlorides". </span>

<span>"In order...to distinguish cation Group I, one adds HCl to a sample. If a Group I cation is present in the sample, a precipitate will form." </span>

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3 years ago

If 100 mL of a 12 M solution of HCl is diluted to a final volume of 500 mL, what would be the final concentration of the diluted

AURORKA [14]

Dilution formula:
mv = MV

where one side is concentration × volume BEFORE dilution and the other side is concentration × volume AFTER dilution.

(100mL) × (12 M) = (500mL) × (X)
(1200 M·mL) = (500mL) × (X)
(1200 M·mL) / (500mL) = X
2.4 M = X

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3 years ago