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Eddi Din [679]
4 years ago
6

Fatima is designing an advanced graphics card for a next-generation video game system. She needs to be able to precisely control

the conductivity of the electronic components. For this purpose, she should use materials that are An element that is likely to have this property is​
Chemistry
2 answers:
polet [3.4K]4 years ago
4 0

Answer:

semiconducting/ tellurium

Explanation:

edgenuity test

belka [17]4 years ago
3 0

Answer:

semiconducting and tellurium

Explanation:

u did the test hope this helps babes

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Could someone explain how they got this answer, explain step by step plz
gulaghasi [49]

Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

Let the abundance of 6–Li be A%

Let the abundance of 7–Li be B%

The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

Atomic mass of Lithium = 6.941amu

The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

Divide both side by 7.42

Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

Therefore, the mass of 6–Li is 6.018 amu

7 0
3 years ago
Is Lilium lancifolium Heterotroph, Autotroph, or Both
Sphinxa [80]
Lilium lancifolium is an autotroph since it is a plant and makes its own food through photosynthesis.
4 0
3 years ago
What is true about ionic compounds?
bezimeni [28]
It’s D I am pretty sure.
6 0
3 years ago
Which mixture is classified as a solution?
svlad2 [7]
Salt water is considered to be a solution
5 0
3 years ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
3 years ago
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