Answer:
Uh first of all this is algebra but I'll answer this
First distribute the three and 5 (Multiply them by both terms inside parenthesis.
3x-6=5x+20
Then add like terms
8x=14
Divide 8 by 8 and 8 by 14
x = 14/8
Explanation:
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
Answer:
FeSO2
Explanation:
Please see attached picture for full solution.
Explanation:
Some Rules Regarding Oxidation Numbers:
- Hydrogen has oxidation number of + 1 except in hydrides where it is -1
- Oxygen has oxidation number of -2 except in peroxides where it is -1
- Some elements have fixed oxidation numbers. E.g Halogen group elements has oxidation number of -1
- Oxidation number of a compound is the sum total of the individual elements and a neutral compound has oxidation number of 0.
A. HI
Hydrogen has oxidation of + 1
Oxidation number of I:
1 + x = 0
x = -1
B. PBr3
Br has oxidation number of - 1
Oxidation number of Pb:
x + 3 (-1) = 0
x = + 3
C. KH
Hydrogen has oxidation of + 1
Oxidation number of K:
1 + x = 0
x = -1
D. H3PO4
Hydrogen has oxidation number of + 1
Oxygen has oxidation number of -2
Oxidation number of P:
3(1) + x + 4(-2) = 0
3 + x - 8 =0
x = 5
<span>My hypothesis is the the cell, having a higher osmolarity than the solution of of nacl in the beaker, will have an osmosis reaction releasing into the solution of nacl. This will continue until both cell and solution reach a balance.</span>
