Answer:
5.6 seconds
Explanation:
The reaction follows a zero-order in dinitrogen monoxide
Rate = k[N20]^0 = change in concentration/time
[N20]^0 = 1
Time = change in concentration of N2O/k
Initial number of moles of N2O = 300 mmol = 300/1000 = 0.3 mol
Initial concentration = moles/volume = 0.3/4 = 0.075
Number of moles after t seconds = 150 mmol = 150/1000 = 0.15 mol
Concentration after t seconds = 0.15/4 = 0.0375 M
Change in concentration of N2O = 0.075 - 0.0375 = 0.0375 M
k = 0.0067 M/s
Time = 0.0375/0.0067 = 5.6 s
The rate of a reaction with this rate law would increase by a factor of 4 if NO concentration were doubled.
One example of a solution<span> is </span>salt water<span> which is a </span>mixture<span> of </span>water<span> and </span>salt<span>. You cannot see </span>the salt<span> and </span>the salt<span> and </span>water<span> will stay a </span>solution<span> if left alone. Parts of a</span>Solution<span>. Solute - The solute is the substance that is being </span>dissolved<span> by another substance.</span>
Answer:
Answers are in the explanation
Explanation:
Ksp of CdF₂ is:
CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)
Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²
When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:
6.44x10⁻³ = [X] [2X]²
6.44x10⁻³ = 4X³
X = 0.1172M
<h3>[F⁻] = 0.2344M</h3><h3 />
Ksp of LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]
When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:
1.84x10⁻³ = [X] [X]
1.84x10⁻³ = X²
X = 0.0429
<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
Each orbit surrounding an atom is allowed A LIMITED NUMBER OF ELECTRONS.
The number of orbit that an atom has is determined by its atomic number, the higher the atomic number the higher the number of orbit in the atom and each orbit has different energy level. Each orbit can only take fixed number of electron. The first shell can only take two electrons while the subsequent shells can only take eight electrons. When an orbit has taken the highest number of electron possible, the remaining electrons are moved to the next shell.
