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aksik [14]
8 months ago
14

How many grams of solute would you use to prepare the following solutions?241.0 mL of 1.11 M NaOHExpress your answer with the ap

propriate units.
Chemistry
1 answer:
Ludmilka [50]8 months ago
8 0
<h2>Answer:</h2>

10.7 grams

<h2>Explanations:</h2>

The formula for calculating the molarity of a solution is given as:

\begin{gathered} molarity=\frac{moles}{volume} \\ moles=molarity\times volume \end{gathered}

Given the following parameters

molarity of NaOH = 1.11M

volume of solution = 241.0mL = 0.241L

Find the moles of NaOH

\begin{gathered} moles\text{ of Na}OH=\frac{1.11mol}{L}\times0.241L \\ moles\text{ of Na}OH=0.2675moles \end{gathered}

Determine the mass of solute (NaOH)

\begin{gathered} Mass\text{ of NaOH}=mole\times molar\text{ mass} \\ Mass\text{ of NaOH}=0.2675\times40 \\ Mass\text{ of NaOH}=10.7grams \end{gathered}

Hence the grams of solute that would be used to prepare the solution is 10.7grams

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A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt
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Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

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integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
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