There are 31.1 grams (g) in 1 troy ounce (ozt). How many ozt of gold are extracted from 1 short ton of average Nevada ore
1 answer:
This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.
<h3>Dimensional analysis</h3>In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.
Thus, since this problem asks for try ounces in an average Nevada ore, which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:
Where the short tons are cancelled out as well as the grams, in order to obtain:
Learn more about dimensional analysis: brainly.com/question/10874167
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Answer:
1) 1,1,1-trichloropropane
2) 1,1,2-trichloropropane
3) 1,2,2-trichloropropane
4) 1,2,3-trichloropropane
Explanation:
For this question, we must remember that isomers are molecules that have the <em>same formula but different structure</em>s. For the formula we can draw a <u>linear chain of three carbons</u> and change the position of the chlorine atoms in the carbon chain.
With this in mind, if we put all the chlorine atoms on the same carbon we will get <u>1,1,1-trichloropropane</u>. If we change an atom from chlorine to carbon 2 we will obtain <u>1,1,2-trichloropropane</u>. If we move another chlorine atom to carbon two we will get <u>1,2,2-trichloropropane</u>. Finally, if we put a chlorine atom in each carbon we will obtain <u>1,2,3-trichloropropane</u>.
See figure one for further explanations
I hope it helps!
