Answer:
Difussion
Explanation:
Diffusion is the result of a totally random phenomenon in which the molecules of a fluid come and go between two vessels that can be connected by a pipe. These molecules travel in a single direction, where the solute is more concentrated to where it is more diluted.
This movement of particles will be modified according to the length or area of the pipe and the concentration of solute. The greater the difference in solute concentration along the tube, the greater the diffusion
Answer:
He had the most potential energy going down hill
Explanation:
because he picked up he went faster
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
<span>Use the Ideal law Equation :
P.V= n.R.T
V = 0.5 L
P = 1.0 atm
</span><span>R= 0.0821 L*atm/mol*K
</span>
<span>n = R*T/P*V
</span><span>P*V= n*R*T
</span>
1.0 * 0.5 = n *<span>0.0821*298
0,5 = n* 24.4658
n = 0,5 / 24.4658
n =0.0204 moles
</span>
Answer:
1. bond in the molecule on the right
Explanation:
CH3CH2-OH
The compound above is an alcohol due to the presence of the OH bond. The wave number of the C - O bond is given as; 1050-1150 cm^-1.
CH3CH__O
The compound above is an aldehyde due to the presence of the CHO bond. The wave number of the C = O bond is given as; 1740-1720 cm^-1
Comparing both bonds, the C = O bond absorbs at a higher wave number.
