Answer:
The electrons farthest from the nucleus of the atom.
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Answer:
Where are the categories, we need an image for reference or something to better understand the question
Answer:
2,375 cans
Explanation:
The strategy here is to use the information given to calculate the lethal dosage contained in the number of cans we will compute.
We know the lethal dosage is
Ld = 10.0 g caffeine
and we also know that the oncentration of caffeine is:
2.85 mg/ oz
So our problem simplifies to calculate how many oz will contain the lethal dose, and then given the ounces per can determine how many cans are required.
First convert the lethal dose in grams to mg:
Ld =( 10 g x 1000 mg ) = 10,000 mg caffeine
10,000 mg x ( 1 Oz / 2.85 mg ) = 28,500 oz
28500 oz x ( 1 can/12 oz ) = 2,375 cans
We could also have calculated it in one step using conversion factors:
Number of cans = 10000 mg x 1 oz/ 2.85 mg x 1 can / oz = 2,375 cans
Answer:
Explanation:
Hello there!
In this case, since the buffer is not given, we assume it is based off ammonia, it means the ammonia-ammonium buffer, whereas the ammonia is the weak base and the ammonium ion stands for the conjugate acid. In such a way, when adding HI to the solution, the base of the buffer, NH3, reacts with the former to promote the following chemical reaction:
Because the HI is totally ionized in solution so the iodide ion becomes an spectator one.
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Answer:
Percentage composition = 14.583%
Explanation:
In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.
Percentage composition by mass of Nitrogen
Nitrogen = 14g/mol
In one mole of the compound;
Mass of Nitrogen = 1 mol * 14g/mol = 14g
Mass of compound = 1 mol * 96.0 g/mol = 96
Percentage composition of Nitrogen = (Mass of Nitrogen / Mass of compound) * 100
percentage composition = 14/96 * 100
Percentage composition = 0.14583 * 100
Percentage composition = 14.583%
