Answer:
NO2
Explanation:
Calculate moles of Nitrogen and oxygen (assume total mass is 100g as you only have percentages therefore there is 63.6g of N and 36.4g of O) This is the ratio that they are in but you need the ratio to be in whole numbers therefore divide each number of moles by 2.275 (the moles of oxygen) to calculate the empirical formula. Hope this helps!
TECTONIC PLATES
a theory explaining the structure of the earth's crust and many associated phenomena as resulting from the interaction of rigid lithospheric plates which move slowly over the underlying mantle.
VALCONES AND TECTONIC
Most volcanoes form at the boundaries of Earth's tectonic plates. These plates are huge slabs of the Earth's crust and upper mantle, which fit together like pieces of a puzzle. These plates are not fixed, but are constantly moving at a very slow rate. They move only a few centimeters per year.
Answer:
term of abc is cde efg hij klm nop
The question is incomplete, here is the complete question:
A bottle of rubbing alcohol having aqueous solution of alcohol contains 70% (v/v) alcohol. If Carl buys a 946 ml bottle of rubbing alcohol, how much of the aqueous solution is water?
<u>Answer:</u> The amount of water present in the given bottle of rubbing alcohol is 283.8 mL
<u>Explanation:</u>
We are given:
Volume of bottle of rubbing alcohol = 946 mL
70% (v/v) alcohol solution
This means that 70 mL of rubbing alcohol is present in 100 mL of solution
Amount of water present in solution = [100 - 70] = 30 mL
Applying unitary method:
In 100 mL of solution, the amount of water present is 30 mL
So, in 946 mL of solution, the amount of water present will be = 
Hence, the amount of water present in the given bottle of rubbing alcohol is 283.8 mL
<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