Answer:
Fourteen hydrogen atoms are bounded in total to the carbon atoms in the structure
Explanation:
The boxes that show "one" indicate that there is only one hydrogen atom bonded to that particular carbon atom while those that show "zero" shows there are no hydrogen atoms bonded to that particular carbon atom. Those that show "three" indicate that there are three hydrogen atoms bonded to that particular carbon.
There are 10 carbon atoms in the structure.
NOTE that each of these carbon atoms must be surrounded with four bonds; which was how the number of hydrogen atoms (numbers in the boxes) weree determined.
The two ways to make a saturated solution are 1 reducing the temperature of the solution and Adding more solute.<span> </span>
It is called rust and it forms when water soaks into the metal forming a chemical reaction.
Answer:
26.5 g
Explanation:
First we convert 100.0 mL to L:
- 100.0 mL / 1000 = 0.100 L
Now we <u>calculate how many moles of sodium carbonate are needed</u>, using the <em>definition of molarity</em>:
- Molarity = moles / liters
- moles = molarity * liters
- 2.5 M * 0.100 L = 0.25 mol
Finally we <u>convert 0.25 moles of sodium carbonate into grams</u>, using its <em>molar mass</em>:
- 0.25 mol * 106 g/mol = 26.5 g
<u>Answer:</u> The cell potential of the cell is +0.118 V
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u> 
<u>Reduction half reaction (cathode):</u> 
In this case, the cathode and anode both are same. So,
will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BNi%5E%7B2%2B%7D_%7Bdiluted%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D_%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
= 
= 1.0 M
Putting values in above equation, we get:


Hence, the cell potential of the cell is +0.118 V