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galben [10]
3 years ago
11

Rob measures the solubility of three different salts in water at 22°C. Which type of graph should he use to display his data?. b

ar graph. line graph. histogram. scatterplot
Chemistry
2 answers:
sveta [45]3 years ago
7 0
To present his data on the <span>solubility of three different salts in water at 22°C, bar graph should be used since there are different salts and the only variable is the type of salt used. Line graph and scatter plot use two coordinates or variables and is common to comparing data using the same sample while using histogram to find out data distribution is irrelevant.</span>
Strike441 [17]3 years ago
7 0
Bar graph should be a perfect choice for Roy for displaying his data. The correct option among all the options that are given in the question is the first option. This is because only a single variable is present and that is the kind of salt that is being used by Rob. I hope the answer has come to your help.
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Butane is burned as a lighter fluid in disposable lighters. Write a balanced equation for the complete oxidation reaction that o
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A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research techni
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Answer:

The pKa of X-281 is 3.73.

Explanation:

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The dissociation equation could be written as follows:

X-281-H ⇆ X-281 + H⁺

Note the equilibrium arrows indicating that not all X-281-H dissociates at equilibrium.

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  X-281-H               ⇆      X-281        +       H⁺

(0.089 M-X)                         X                    X

We also know that the pH is 2.40. Then:

pH = -log[H⁺] = 2.40

where [H⁺] is the molar concentration of the protons or "X". Then:

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X = 10^(-2.40) = 3.98 x 10⁻³ M

The concentration of each species present in the equilibrium is then:

[H⁺] = 3.98 x 10⁻³ M

[X-281] = 3.98 x 10⁻³ M

[X-281-H] = 0.089 M - 3.98 x 10⁻³M = 0.085 M

At equlibrium, the acidity constant Ka is:

Ka = [X-281] * [H⁺] / [X-281-H]

Ka = (3.98 x 10⁻³ M * 3.98 x 10⁻³ M ) / 0.085 M = 1.86 x 10⁻⁴

Then the pKa is:

pKa = -log Ka = -log (1.86 x 10⁻⁴) = <u>3.73</u>

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