<u>Answer:</u> The volume of concentrated solution required is 9.95 mL
<u>Explanation:</u>
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
pH = 0.70
Putting values in above equation, we get:
![0.70=-\log[H^+]](https://tex.z-dn.net/?f=0.70%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-0.70}=0.199M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-0.70%7D%3D0.199M)
1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.
Molarity of nitric acid = 0.199 M
To calculate the volume of the concentrated solution, we use the equation:
where,
are the molarity and volume of the concentrated nitric acid solution
are the molarity and volume of diluted nitric acid solution
We are given:
Putting values in above equation, we get:
Hence, the volume of concentrated solution required is 9.95 mL
Answer:
10.3 g of oxygen are formed when 26.4 g of potassium chlorate is heated
Explanation:
This is the balanced equation:
2KClO₃(s) → 2KCl(s) + 3O₂(g)
Ratio beteween the salt and oxygen is 2:3
Molar mass of KClO₃ = 122.55 g/m
Let's find out the moles of salt
Mass / Molar mass
26.4 g /122.55 g/m = 0.215 moles
So, this is the final rule of three:
If 2 moles of KClO₃ make 3 moles of oxygen
0.215 moles of KClO₃ make (0.215 .3) /2 = 0.323 moles of O₂ are produced
Molar mass O₂ = 32 g/m
Moles . molar mass = mass
0.323 m . 32g/m = 10.3 g
Answer:
[H₃O⁺] = 0.05 M & [OH⁻] = 2.0 x 10⁻¹³.
Explanation:
- HNO₃ is completely ionized in water as:
<em>HNO₃ + H₂O → H₃O⁺ + NO₃⁻.</em>
- The concentration of hydronium ion is equal to the concentration of HNO₃:
[H₃O⁺] = 0.05 M.
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] </em>= 10⁻¹⁴/0.05 = <em>2.0 x 10⁻¹³.</em>
Answer:
Explanation:
a. CuO+ 2HCl⇒CuCl2+ H2O
b. 
= 
= 0,05 (mol)
⇒
= =0,05 mol
⇒
= 0,05×135=6,75 (g)
c. 
=2× =0,1 (mol)
⇒
= 0,1×36,5= 3,65 (g)
⇒
= 
×100=36,5 (g)
⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=
×100=
×100≈ 16,67%
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