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devlian [24]
2 years ago
9

Suppose you work at a theme park. Your supervisor wants you to make a sign displaying the maximum weight that a roller coaster t

rain can carry. Your supervisor knows that the maximum weight is 1686.5 kg. However, he wants the sign to be quickly understood and tells you to make a sign that says: Maximum Weight 1700 kg. How could the lack of precision in this example cause problems?
Chemistry
1 answer:
grin007 [14]2 years ago
3 0

Answer:

amusement parks. Each day, we flock by the millions to the nearest park, paying a sizable hunk of money to wait in long lines for a short 60-second ride on our favorite roller coaster. The thought prompts one to consider what is it about a roller coaster ride that provides such widespread excitement among so many of us and such dreadful fear in the rest? Is our excitement about coasters due to their high speeds? Absolutely not! In fact, it would be foolish to spend so much time and money to ride a selection of roller coasters if it were for reasons of speed. It is more than likely that most of us sustain higher speeds on our ride along the interstate highway on the way to the amusement park than we do once we enter the park. The thrill of roller coasters is not due to their speed, but rather due to their accelerations and to the feelings of weightlessness and weightiness that they produce. Roller coasters thrill us because of their ability to accelerate us downward one moment and upwards the next; leftwards one moment and rightwards the next. Roller coasters are about acceleration; that's what makes them thrilling. And in this part of Lesson 2, we will focus on the centripetal acceleration experienced by riders within the circular-shaped sections of a roller coaster track. These sections include the clothoid loops (that we will approximate as a circle), the sharp 180-degree banked turns, and the small dips and hills found along otherwise straight sections of the track.

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2 years ago
An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density
V125BC [204]

Answer:

Au

Explanation:

For the density of a face-centered cubic:

Density = \dfrac{4 \times M_w}{N_A \times a^3}

where

M_w = molar mass of the compound

N_A= avogadro's constant

a^3 = the volume of a unit cell

Given that:

Density (\rho) = 19.30 g/cm³

a = 0.408 nm

a = 0.408 \times 10^{-9} \times 10^{2} \ cm

a = 4.08 \times 10^ {-8} \ cm

∴

19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}

M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}

M_w=197.37 \ g/mol

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The ka values for several weak acids are given below. which acid (and its conjugate base) would be the best buffer at ph = 8.0?
Pie
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To support this answer, we first calculate for the pKa value as the negative logarithm of the Ka value: 
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For Tris, which is an abbreviation for 2-Amino-2-hydroxymethyl-propane-1,3 -diol and has a Ka value of 6.3 x 10^-9, the pKa is
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We know that buffers work best when pH is equal to pKa:
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Therefore Tris would be a best buffer at pH = 8.0.
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