Answer: assume pathogens are present and treat the samples accordingly
Explanation:
When investigators are unable to conclusively ascertain the source of a biological sample found at a crime scene, the correct thing to do is to treat it as if pathogens are present in it and handle it according to set rules on how to handle pathogens.
This is done to ensure that if a pathogen is indeed present, it would not cause a health emergency by infecting those who come in contact with the samples at the scene.
Your Answer Will Be Intensive Property
Answer:
25 grams
Explanation:
You strat off with 400 grams of your substance. By day 14, half has dcayed and you only have 200 grams left. By day 28, there are 100 grams of the substance. On day 42, there are 50 grams left. Finally, on day 56, the substance has been through four half-lives and 25 grams remain.
Answer:
.
Explanation:
Electrons are conserved in a chemical equation.
The superscript of indicates that each of these ions carries a charge of . That corresponds to the shortage of one electron for each ion.
Similarly, the superscript on each ion indicates a shortage of three electrons per such ion.
Assume that the coefficient of (among the reactants) is , and that the coefficient of (among the reactants) is .
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There would thus be silver () atoms and aluminum () atoms on either side of the equation. Hence, the coefficient for and would be and , respectively.
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The ions on the left-hand side of the equation would correspond to the shortage of electrons. On the other hand, the ions on the right-hand side of this equation would correspond to the shortage of electrons.
Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of electrons, the right-hand side should also be electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of electrons. These two expressions should have the same value. Therefore, .
The smallest integer and that could satisfy this relation are and . The equation becomes:
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