<span>The number of electrons in an atom's outermost valence shell governs its bonding behavior.
In N</span>₂, three electrons are being shared by each nitrogen atom, making a total of 6 shared electrons.
In CCl₄, 4 electrons are being shared by each carbon atom and 1 electron is being shared by each chlorine atom
In SiO₂, 4 electrons are being shared by each silicon atom and 2 electrons are being shared by each oxygen atom.
In AlCl₃, 3 electrons are being shared by each aluminum atom and 1 electron is being shared by each Cl atom
In CaCl₂, 2 electrons are lost by the calcium atom and 1 electron is gained by each chlorine atom
In LiBr, 1 electron is lost by the lithium atom and 1 electron is gained by the bromine atom
Answer:
6.142 moles of NaCl
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2AlCl3 + 3Na2S —> Al2S3 + 6NaCl
Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:
Mass mass of Na2S = 78.048g/mol
Mass of Na2S = 239.7g
Number of mole Na2S =..?
Mole = Mass /Molar Mass
Number of mole Na2S = 239.7/78.048 = 3.071 moles
Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:
From the balanced equation above,
3 moles of Na2S reacted to produce 6 moles of NaCl.
Therefore, 3.071 moles of Na2S will react to produce = (3.071 x 6)/3 = 6.142 moles of NaCl
Answer:
Oxygen and sulfur have similar chemical properties because both elements have six electrons in their outermost electron shells. Indeed, both oxygen and sulfur form molecules with two hydrogen atoms: water and hydrogen sulfide
Explanation:
Answer:
37 mmol of acetate need to add to this solution.
Explanation:
Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-
![pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%28acetic%20acid%29%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7Bmmol%20of%20CH_%7B3%7DCOOH%20%7D%5D)
Here pH is 5.31, 
(acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.
Plug in all the values in the above equation:
![5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]](https://tex.z-dn.net/?f=5.31%3D4.74%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7B10%7D%5D)
or, mmol of 
= 37
So 37 mmol of acetate need to add to this solution.
