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Eddi Din [679]
3 years ago
15

What would you need to know in order to calculate how many pre-1982 pennies and post-1982 pennies are in the mixed sample?

Chemistry
1 answer:
ololo11 [35]3 years ago
8 0
<span>In 1982, the physical composition of the penny was changed; the ratio of the zinc to copper was essentially reversed. As a result, a pre-82 penny has a different density from a post-82 penny. It would help to know several things: first, the average mass and density of both a single pre and post 82 penny. Then you need to know how many pennies in total are in the mixed sample. You would also need to know the mass and density measures of "pure" samples, that is, if we have a mixed sample of 20 pennies, it would be useful to know the mass and density measurements of 20 pre-82 and 20 post-82 pennies. You can then figure out your estimate based on the variances in the density measurements.</span>
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<span>So left handed approximation underestimates the area under a increasing curve and over estimates for decreasing curves. And right handed approximation overestimates for increasing curves and underestimates for decreasing curves.</span>
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3 years ago
Computers have the mineral _______________ in their motherboard.
lana66690 [7]

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silica

Explanation:

6 0
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How did Wilma contribute to change in America
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She overcame her disabilities to compete in the 1956 Summer Olympic Games, and in 1960, she became the first American woman to win three gold medals in track and field at a single Olympics. Later in life, she formed the Wilma Rudolph Foundation to promote amateur athletics.

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3 years ago
What mass of magnesium would combine with exactly 16.0 grams of oxygen
TEA [102]
1.65g MgO = 1g Mg
1.65 - 1 = 0.65 g of O in MgO

solve it using proportion:
1g Mg / 0.65g O = x (g) Mg / 16g O
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24.6 g is the answer.

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3 0
3 years ago
An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions at room
Andrej [43]

Answer:

0.758 V.

Explanation:

Hello!

In this case, case when we include the effect of concentration on an electrochemical cell, we need to consider the Nerst equation at 25 °C:

E=E\°-\frac{0.0591}{n} log(Q)

Whereas n stands for the number of moles of transferred electrons and Q the reaction quotient relating the concentration of the oxidized species over the concentration of the reduced species. In such a way, we can write the undergoing half-reactions in the cell, considering the iron's one is reversed because it has the most positive standard potential so it tends to reduction:

Fe^{2+}+2e^-\rightarrow Fe^0\ \ \ E\°=0.440V\\\\Ni^0\rightarrow Ni^{2+}+2e^-\ \ \ E\°=-0.250V

It means that the concentration of the oxidized species is 0.002 M (that of nickel), that of the reduced species is 0.40 M and there are two moles of transferred electrons; therefore, the generated potential turns out:

E=(0.440V+0.250V)-\frac{0.0591}{2} log(\frac{0.002M}{0.40M} )\\\\E=0.758V

Beat regards!

8 0
3 years ago
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