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motikmotik
2 years ago
8

Questions & Conclusions:

Chemistry
1 answer:
dem82 [27]2 years ago
3 0

The amount of KNO₃ precipitated out of solution when you cooled the solution from 100°C to -22°C is obtained from the solubility curve

<h3>What is a solubility curve?</h3>

A solubility curve is a curve of the solubility of a solute against temperature.

The solubility curve shows that the solubility of different solute at different temperatures.

The solubility curve of KNO₃ is as shown. Solubility at 100 °C and  -22 °C is not shown in the curve.

However, the amount of KNO₃ precipitated out of solution when cooled the solution from 100 °C to -22 °C can be determined by subtracting the amount of solute dissolve at -22 °C from that dissolved at 100 °C.

In conclusion, the solubility curve is used to determined the amount of solute dissolved in a given volume of solvent at different temperatures.

Learn more about solubility curve at: brainly.com/question/928930

#SPJ1

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The reaction (CH3)3CBr + OH- (CH3)3COH + Br- in a certain solvent is first order with respect to (CH3)3CBr and zero order with r
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Answer and Explanation:

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

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Comparing this to the equation of a straight line; y = mx + c

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b) c = In A = 33.5

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c) K = A(e^(-Ea/RT))

A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

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2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'
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Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

\displaystyle M_1V_1= M_2V_2

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}

Convert this value to mL:
\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

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