Answer:
<em>A process that involves rearrangement of the molecular or ionic structure of a substance, as opposed to a change in</em> <u><em>physical form or a nuclear reaction.</em></u>
Explanation:
A process that involves rearrangement of the molecular or ionic structure of a substance, as opposed to a change in <u><em>physical form or a nuclear reaction.</em></u>
Answer: The energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.
Explanation:
Given: n = 2
The relation between energy and 
orbit of an atom is as follows.
Substitute the values into above formula as follows.
The negative sign indicates that energy is being released.
Thus, we can conclude that the energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.
Fossil C is the youngest. Layers further down were placed earlier than others. Because C is on the top layer, it must be the youngest (unless a disaster mixed up all the layers!).
Answer:
OptionA. 2C4H10 + 13O2 —> 8CO2 + 10H20
Explanation:
Butane burns is air (O2) according to the equation:
C4H10 + O2 —> CO2 + H20
Considering the equation, it is evident that it not balanced. Now let us balance the equation as shown below;
There are a total of 4 carbon atoms on the left and 1 carbon atom on the right. It can be balanced by putting 4 in front of CO2 as shown below:
C4H10 + O2 —> 4CO2 + H20
Next, there are 10 hydrogen atoms on the left and 2 hydrogen atoms on the right. Therefore to balance it, put 5 in front of H2O as shown below:
C4H10 + O2 —> 4CO2 + 5H20
Now, there are a total of 13 oxygen atoms on the right and 2 at the left. To balance it, put 13/2 in front of O2
as shown below
C4H10 + 13/2O2 —> 4CO2 + 5H20
Now we multiply through by 2 clear off the fraction and we obtained:
2C4H10 + 13O2 —> 8CO2 + 10H20
As per the law of constant composition, a given sample will always contain the same number of elements that combine in the same mass proportion.
Therefore if a sample of 13.97 g of NaBr contains 22.39 % of Na by mass then, a sample of 5.75 g of NaBr would also contain 22.39% Na by mass
Hence:
Mass of Na = 5.75 g * 22.39/100 = 1.287 g
5.75 g of NaBr would contain 1.29 g of Na
