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3 years ago

How many atoms are in .47 moles of argon

Chemistry

1 answer:

lesantik [10]3 years ago

3 0

Answer:

$\large \boxed {2.8 \times 10^{23}\text{ atoms Ar}}$

Explanation:

$\text{Atoms of Ar} = \text{0.47 mol Ar} \times \dfrac{ 6.022 \times 10^{23}\text{ atoms Ar}}{\text{1 mol Ar}}\\\\= \large \boxed {\mathbf{2.8 \times 10^{23}}\textbf{ atoms Ar}}$

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4 years ago

Atypicalaspirintabletcontains325mgofacetylsalicylic acid (HC9H7O4). Calculate the pH of a solution that is prepared by dissolvin

Sergio [31]

Answer:

$\boxed{2.65}$

Explanation:

1. Mass of acetylsalicylic acid (ASA)

$m = \text{2 tablets} \times \dfrac{\text{325 mg}}{\text{1 tablet}} = \text{750 mg}$

2. Moles of ASA

HC₉H₇O₄ =180.16 g/mol

$n = \text{750 mg} \times \dfrac{\text{1 mmol}}{\text{180.16 mg }} = \text{4.163 mmol}$

3. Concentration of ASA

$c = \dfrac{\text{4.163 mmol}}{\text{237 mL}} = \text{0.01757 mol/L}$

4. Set up an ICE table

$\begin{array}{ccccccc}\text{HA} & + & \text{H$_{2}$O}& \, \rightleftharpoons \, &\text{H$_{3}$O$^{+}$} & + &\text{A}^{-}\\0.01757 & & & &0 & & 0 \\-x & & & &+x & & +x \\0.01757-x & & & &x & & x \\\end{array}\\$

5. Solve for x

$K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}]\text{A}^{-}]} {\text{[HA]}} = 3.33 \times 10^{-4}\\\\\dfrac{x^{2}}{0.01757 - x} = 3.33 \times 10^{-4}\\\\\textbf{Check that }\mathbf{x \ll 0.01757}\\\\\dfrac{ 0.01757 }{3.33 \times 10^{-4}} = 53 < 400\\\\\text{The ratio is less than 400. We must solve a quadratic equation.}\\\\x^{2} = 3.33 \times 10^{-4}(0.01757 - x) \\\\x^{2} = 5.851 \times 10^{-6} - 3.33 \times 10^{-4}x\\\\x^{2} + 3.33 \times 10^{-4}x - 5.851 \times 10^{-6} = 0$

6. Solve the quadratic equation.

$a = 1; b = 3.33 \times 10^{-4}; c = -5.851 \times 10^{-6}$

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{Substituting values into the formula, we get}\\x = 0.002258\qquad x = -0.002591\\\text{We reject the negative value, so}\\x = 0.002258$

7. Calculate the pH

$\rm [H_{3}O^{+}]= x \, mol \cdot L^{-1} = 0.002258 \, mol \cdot L^{-1}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.002258} = \mathbf{2.65}\\\text{The pH of the solution is } \boxed{\textbf{2.65}}$