When water chemically combines with carbon dioxide, a Carbonic acid is formed.
<u>Explanation</u>:
- Carbon dioxide responds with water in a solution to form a weak acid, carbonic acid. Carbonic acid disassociates into hydrogen particles and bicarbonate particles. The hydrogen particles and water respond with the most basic minerals modifying the minerals.
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Carbon dioxide and the other atmospheric gases disintegrate in surface waters. Dissolved gases are in equilibrium with the gas in the atmosphere. Carbon dioxide responds with water in a solution to form the weak acid, carbonic acid. Carbonic acid disassociates into hydrogen particles and bicarbonate particles.
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The hydrogen particles and water respond with the most basic minerals altering the minerals. The results of enduring are prevalently clays and soluble particles, for example, calcium, iron, sodium, and potassium. Bicarbonate particles additionally remain in solution; a remnant of the carbonic acid that was utilized to weather the rocks.
Uses of nonmetals in our daily life: Oxygen which is 21% by volume helps in the respiration process.
Nonmetals used in fertilizers: Fertilizers contain nitrogen.
Nonmetals used in crackers: Sulphur and phosphorus are used in fireworks.
The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution
there are 1.00 mol in 1 L of solution
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L
volume of HCl required is 0.150 L
<u>Answer:</u> The equilibrium concentration of water is 0.597 M
<u>Explanation:</u>
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:
The expression for 
is written as:
![K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
The concentration of pure solids and pure liquids are taken as 1 in the expression.
For the given chemical reaction:
The expression of 
for above equation is:
![K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2S%5D%5E2%5Ctimes%20%5BO_2%5D%7D)
We are given:
![[H_2S]_{eq}=0.671M](https://tex.z-dn.net/?f=%5BH_2S%5D_%7Beq%7D%3D0.671M)
![[O_2]_{eq}=0.587M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D0.587M)
Putting values in above expression, we get:
![1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}](https://tex.z-dn.net/?f=1.35%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%280.671%29%5E2%5Ctimes%200.587%7D)
![[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Csqrt%7B%281.35%5Ctimes%200.671%5Ctimes%200.671%5Ctimes%200.587%29%7D%3D0.597M)
Hence, the equilibrium concentration of water is 0.597 M
