Answer:
b. The number of electrons
Explanation:
A "neutral atom" has a <u>neutral charge</u>. This means that <em>its charge is equal to </em><em>zero. </em>In order for the charges to cancel out each other, the atom's <em>positive charge should be equal to the negative charge. </em>These being said, the number of electrons<em> (negatively-charged)</em> is then equal to the number of protons <em>(positively-charged). </em>Those atoms which are not neutral are called <em>"ions."</em> This means that they either have more or less electrons than the protons.
The answer is 18.
This is obtained by multiplying the three oxygen atoms in NO₃, that are inside the parenthesis by the number two outside the parenthesis. The number 2 is there because it comes from associating with Fe ²⁺ .
3×2=6 oxygen atoms present in the formula.
Then multiplied by the 3 that's in front of the formula (that indicates how many molecules).
6 x 3= 18
The percentage of elements in the compound is determined from the number of elements in the compound
<h3>Further explanation </h3>
Proust states the Comparative Law that compounds are formed from elements with the same Mass Comparison so that compounds have a fixed composition of elements
![\tt \%A~in~A_xB_y=\dfrac{x\times Ar~A}{MW~A_xB_y}\times 100\%](https://tex.z-dn.net/?f=%5Ctt%20%5C%25A~in~A_xB_y%3D%5Cdfrac%7Bx%5Ctimes%20Ar~A%7D%7BMW~A_xB_y%7D%5Ctimes%20100%5C%25)
![\tt \%B~in~A_xB_y=\dfrac{y\times Ar~B}{MW~A_xB_y}\times 100\%](https://tex.z-dn.net/?f=%5Ctt%20%5C%25B~in~A_xB_y%3D%5Cdfrac%7By%5Ctimes%20Ar~B%7D%7BMW~A_xB_y%7D%5Ctimes%20100%5C%25)
Example
% Lithium (Li) in Li₂CO₃(lithium carbonate)
Ar of Li = 6.941 g/mol
MW of Li₂CO₃ = 73.891 g/mol
![\tt \%Li=\dfrac{2.Ar~Li}{MW~Li_2CO_3}\times 100\%\\\\\%Li=\dfrac{2\times 6.941}{73.891}\times 100\%=18.79\%](https://tex.z-dn.net/?f=%5Ctt%20%5C%25Li%3D%5Cdfrac%7B2.Ar~Li%7D%7BMW~Li_2CO_3%7D%5Ctimes%20100%5C%25%5C%5C%5C%5C%5C%25Li%3D%5Cdfrac%7B2%5Ctimes%206.941%7D%7B73.891%7D%5Ctimes%20100%5C%25%3D18.79%5C%25)
Answer:
x = 33.52 amu
Explanation:
It is given that,
Isotope A has a mass of 34 amu and an abundance of 52%, isotope B has a mass of 33 amu and an abundance of 48%.
Let x is the average atomic mass of this element. It can be calculated as follows :
![x=52\%\ \text{of}\ 34+48\%\ \text{of}\ 33\\\\x=\dfrac{52}{100}\times 34+\dfrac{48}{100}\times 33\\\\x=0.52\times 34+0.48\times 33\\\\x=33.52\ \text{amu}](https://tex.z-dn.net/?f=x%3D52%5C%25%5C%20%5Ctext%7Bof%7D%5C%2034%2B48%5C%25%5C%20%5Ctext%7Bof%7D%5C%2033%5C%5C%5C%5Cx%3D%5Cdfrac%7B52%7D%7B100%7D%5Ctimes%2034%2B%5Cdfrac%7B48%7D%7B100%7D%5Ctimes%2033%5C%5C%5C%5Cx%3D0.52%5Ctimes%2034%2B0.48%5Ctimes%2033%5C%5C%5C%5Cx%3D33.52%5C%20%5Ctext%7Bamu%7D)
So, the average atomic mass of this element is 33.52 amu.
Answer:
Dissolved oxygen and temperature
Explanation:
According to study.com:
<em>"For aquatic ecosystems, these factors include light levels, water flow rate, </em><em>temperature</em><em>, </em><em>dissolved oxygen</em><em>, acidity (pH), salinity and depth. "</em>
According to ck12.org:
<em>"Aquatic biomes are distinguished by the availability of sunlight and the concentration of </em><em>dissolved oxygen</em><em> and nutrients in the water."</em>