Answer:
S= 2(1) = 2
O= 2(4) = 8
Na= 2(2) = 4
Explanation:
The given compound is:
2Na₂SO₄
An element is a distinct substance that cannot be split up into simpler substances.
So;
Number of atoms of elements here are:
S= 2(1) = 2
O= 2(4) = 8
Na= 2(2) = 4
Answer:
carbon mass = 12.01g/mol
hydrogen mass = 1.01g/mol
4 carbon atoms and 10 hydrogen so
12.01 x 4 + 1.01 x 10
48.04g/mol + 10.10g/mol
= 58.14g/mol
Answer:
11.25moles of CO2
Explanation:
First, let us generate a balanced equation for the reaction of propane to produce CO2. This reaction called Combustion. It is a reaction in which propane burns in air (O2) to produce CO2 and H20. The equation is given below:
C3H8 + 5O2 —> 3CO2 + 4H2O
From the equation,
1mole of C3H8 produced 3moles of CO2.
Therefore, 3.750 moles of C3H8 will produce = 3.750 x 3 = 11.25moles of CO2
1. Find its coordination figure/coordination number of central atom (CF)
Ev = Vallence electron of central atom
Σe = electrons donated from substituents
Terminal O gives 0 electrons, hence Σe = 3 x 0
charge = charge of the compound
2. Find EP (electron pairs) and LP (lone pairs)
LP = CF - EP
3. Draw the skeleton with octet substituents (top right figure)
4. Find formal charge for each atoms (Qf)
5. Write formal charge near atom in skeleton
6. Enjoy
"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.
Option: b
<u>Explanation</u>:
As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant
for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.
To calculate
firstly molarity of ions are needed to be found with formula: 
Then at equilibrium cations and anions concentration is considered same hence:
![\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathbf%7BA%7D%20%5Cmathbf%7Bg%7D%5E%7B%2B%7D%5Cright%5D%3D%5B%5Cmathbf%7BC%7D%20%5Cmathbf%7BI%7D%5D%3D%5Ctext%20%7B%20molarity%20of%20ions%20%7D)
Hence from above data
can be calculated by:
= ![\left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathbf%7BA%7D%20%5Cmathbf%7Bg%7D%5E%7B%2B%7D%5Cright%5D%20%5Ccdot%5B%5Cmathbf%7BC%7D%20%5Cmathbf%7BI%7D%5D)