Chemistry, 03.05.2020 14:26, Day1201

What are the pros and cons of modern medicine? (I'm rewarding 25 points)

Answers: 2

Show answers

Answers

Pros: The rate at which modern medicine is advancing is astonishing. You can sit on your couch and watch TV and see commercials for new medications at an alarming pace — and these new medications have reduced our death rates for stroke, heart disease, and cancer. In fact, due to modern medicine, the death rate from heart disease has decreased by 60% since 1970! The death rate for HIV/AIDS has dropped more than 75% since 1995, when its was at its peak.

Cons: However, critics of modern medicine are quick to point out that there is a pill for everything . Yes we are quick to create more drugs, and yes, we are living longer due to said drugs but are we living better?
Medicine has evolved, and we now have surgeons who are able to perform surgeries robotically. But critics point out that medicine has evolved to the point that we are treating disease instead of preventing it .

answered: Guest

The correct answer is weight.

answered: Guest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.