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4 years ago

Which of the following is an element 1 .carbon dioxide 2. sodium 3. Ammonia 4. sand

Chemistry

2 answers:

Leto [7]4 years ago

8 0

Answer is sodium, because the rest is a compound & sand is a mixture.

posledela4 years ago

8 0

Answer: 2) "sodium" .
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(or, " Na " ; as it appears on the Periodic Table of Elements).
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Note:
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Answer choice: 1) carbon dioxide ; or, " CO₂ " ; is a compound of TWO elements.
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Answer choice: 3) ammonia ; or, " NH₃ " ; is a compound of TWO or more elements; and is often found in a solution, or "mixture.
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Answer choice: 4) sand — variety of types — is a mixture of more than one elements; or most often, either: 1) a mixture of compounds; or 2) a mixture of various compounds and elements.
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Answer choice: 2) sodium ; or, " Na " ; is a "single element".
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Answer:

$\boxed{2.65}$

Explanation:

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

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$\begin{array}{ccccccc}\text{HA} & + & \text{H$_{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

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

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