Describe one benefit and one drawback of using models to represent scientific processes?

1 answer:

3 0

<span>Among the benefits; enlarging so that you can see it visually, making something complex and hard to understand to simplify it. One draw back is that not entirely accurate because it is basic. Additionally; models help us to visualize smaller structures that are too small to see properly. However; they do not take all variables into account and thus they may be inaccurate. </span>

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This happens in diffusion. Choose the correct answer(s)

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The answer is d. particles move from an area of high concentration to low concentration.

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In 1887, Albert A. Michelson and Edward W. Morley conducted an experiment to measure the speed at which the Earth moves through

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Read 2 more answers

If an equal number of moles of the weak acid HF and the strong base KOH are added to water, is the resulting solution acidic, ba

s344n2d4d5 [400]

Answer:

Being a weak acid and a strong base, where it is diluted in a neutral medium such as water, the basic medium predominates, almost alkaline pH.

Explanation:

The acidity of the solution, being weak, means that its pH is not so low, therefore it will be easier to reach the values of 7 or 7 where alkalinity or basity is indicated.

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500.0 mL of a gas is collected at 745.0 mm Hg. what what will the volume be at standard pressure

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This is an application of Boyle's law:
P₁V₁ = P₂V₂. we don't have to convert volume and pressure to standard forms. we can even use the pressure with mmHg
1 atm = 760 mmHg
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What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}$

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