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kramer
3 years ago
14

What does the interquartile range represent?

Chemistry
2 answers:
valkas [14]3 years ago
7 0

Answer:

c

Explanation:

Marizza181 [45]3 years ago
3 0

Answer: C) middle 50 percent of the data

The interquartile range (IQR) spans from the first quartile Q1 to the third quartile Q3.

25% of the data is below Q1 and 75% of the data is below Q3. The gap between the two endpoints consists of 75-25 = 50 percent of the data, or half of the data.

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The density of NaCl( s) is 2.165 g cm 3 at 25 C. How will the solubility of NaCl in water be affected by an increase in pressure
sammy [17]

The solubility of NaCl in water will not be affected by an increase in pressure.

We know that the density of NaCl(s) in 2.165 g/cm³ at 25 °C and we want to know how will its solubility in water be affected when the pressure is increased.

<h3>What is solubility?</h3>

Solubility is the maximum mass of a solute that can be dissolved in 100 grams of solvent at a determined temperature.

The solubility of a solid, such as NaCl, in a liquid, is mainly affected by the temperature. However, since solids are not compressible, an increase in pressure will not affect its solubility.

On the other hand, the solubility of gases in water will increase with an increase in pressure, as stated by Henry's law.

The solubility of NaCl in water will not be affected by an increase in pressure.

Learn more about solubility here: brainly.com/question/11963573

7 0
2 years ago
A family pool holds 10,000 gallons of water. How many cubic centimeters is this?
Ierofanga [76]

Answer:

3.785412e+7cm³

Explanation:

Have a nice day.

3 0
2 years ago
How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward
fomenos

Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

NH3 + H2O(l) ⇄ NH4+ + OH-

Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

<em>[NH₄⁺] = [OH⁻] = X</em>

<em>And as </em>[NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

pH:

As pOH = -log [OH-]

pOH = 2.48

pH = 14 - pOH

<h3>pH = 11.52</h3>
5 0
2 years ago
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