radio waves, micro waves, infrared waves, visible light rays, ultra-violet rays, x-rays, then finally, gamma rays. I hope this was helpful to you.
Answer:
18,1 mL of a 0,304M HCl solution.
Explanation:
The neutralization reaction of Ba(OH)₂ with HCl is:
2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O
The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:
= 2,7531x10⁻³moles of Ba(OH)₂
By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:
= 5,5062x10⁻³moles of HCl.
The volume of a 0,304M HCl solution for a complete neutralization is:
= 0,0181L≡18,1mL
I hope it helps!
POH value was calculated by the negative logarithm of hydroxide ion concentration.
To know the hydrogen ion concentration, we need to know the pH value, that can be found out if pOH is known
pH + pOH = 14
pH = 14 - pOH
pH = 10.65
once the pH is known we have to find the antilog.
[H⁺] = antilog (-pH)
antilog can be found by
[H⁺] = 10^(-10.65)
[H⁺] = 2.2 x 10⁻¹¹ M
It is more slippery, and it is heavier
Answer:
[Ag⁺] = 5.0x10⁻¹⁴M
Explanation:
The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:
Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷
The PbI₂ <em>just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>
<em />
As the initial [Pb²⁺] = 0.0050M:
[Pb²⁺] [I⁻]² = 1.4x10⁻⁸
[0.0050] [I⁻]² = 1.4x10⁻⁸
[I⁻]² = 1.4x10⁻⁸ / 0.0050
[I⁻]² = 2.8x10⁻⁶
<h3>[I⁻] = 1.67x10⁻³</h3><h3 />
So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:
[Ag⁺] [I⁻] = 8.3x10⁻¹⁷
[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷
<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>