Answer:
NaCi + LiCl is the correct answer
Answer:
Decreasing order of strength of the the acids :
Explanation:
The strength of an acid is measured by their pH of aqueous solution.
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![pH\propto \frac{1}{[H^+]}](https://tex.z-dn.net/?f=pH%5Cpropto%20%5Cfrac%7B1%7D%7B%5BH%5E%2B%5D%7D)
- Lower the pH more will be the hydrogen ions and stronger will be the acid.
- Higher the pH less will be the hydrogen ions and weaker will be the acid.
Solution of HX , has equal number of hydrogen ions as a that of its initial molecules o HX.
![[H^+]_x=[HX]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_x%3D%5BHX%5D)
Solution of HY , molecules of hydrogen ions are half of the molecules of HY.
![[H^+]_y=\frac{1}{2}[HY]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_y%3D%5Cfrac%7B1%7D%7B2%7D%5BHY%5D)
Solution of HZ , one quarter molecules of hydrogen ions and three quarter of the molecules of HY.
![[H^+]_z=\frac{1}{4}[HZ]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_z%3D%5Cfrac%7B1%7D%7B4%7D%5BHZ%5D)
![[H^+]_x>[H^+]_y>[H^+]_z](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_x%3E%5BH%5E%2B%5D_y%3E%5BH%5E%2B%5D_z)
This means that HX is strongest acid followed by HY and then HZ.
Decreasing order of strength of the the acids :
Answer:
The answer to your question is letter B. 2
Explanation:
Unbalanced reaction
NH₃ + O₂ ⇒ HNO₃ + H₂O
Reactants Elements Products
1 N 1
3 H 3
2 O 4
Balanced reaction
NH₃ + 2O₂ ⇒ HNO₃ + H₂O
Reactants Elements Products
1 N 1
3 H 3
4 O 4
When the reaction is completely balanced the coefficient of O₂ is 2
a.
Acids react with bases and give salt and water and the products.
Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
To balance the reaction equation, both sides hould have same number of elements.
Left hand side, Right hand side,
H atoms = 2 H atoms = 2
Cl atoms = 1 Cl atoms = 1
Na atoms = 1 Na atoms = 1
O atoms = 1 O atoms = 1
Hence, the reaction equation is already balanced.
b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
= 5.681 x 10⁻³ mol
Stoichiometric ratio between NaOH and HCl is 1 : 1
Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol
5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.
Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
= 0.21 M
