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12345 [234]
3 years ago
8

I will give brainliest! I have an assignment about career exploration. I have to create a 5 minute news segment on why the schoo

l should not get rid of chemistry class. A question I have to include is, "What opportunities can a chemist, chemical engineer, or molecular biologist able to take advantage of as a result of their background in chemistry?"
Chemistry
1 answer:
andriy [413]3 years ago
5 0

Answer:

There are many opportunities a chemist, chemical engineer, and molecular biologist may have. These professions and or practices could become Product Development Scientists, a Clinical Pharmacist, a Biochemist, or a Pharmaceutical Administer. These careers are very important, because people with professions in these areas will be able to get well paying jobs that will most likely have increasing value over time.

Schools need to offer chemistry so people seeking jobs in these areas will be able to do what they love, and for those who are seeking other science related professions like becoming nurses and other important members in the work force, will be able to get the education they need. Chemistry is a college level class, and is a college credit. If you're not seeking a higher profession in the scientific area, you can take it to get a required credit out of the way.

There are a plethora of science jobs out there that are crucial to the wellbeing of people, and these jobs will ALWAYS be important.

This isn't five minutes... but I hope this helps a little. Add a lil' bit more, and you should be good. I really hope this helped.

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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}}

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m(H₂)=M(H₂)vc/2

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1.2 grams of hydrogen gas will be produced
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