The best answer among the choices listed is the third option. The rate over the course of a reaction <span>decreases because the concentration of reactants decreases. As the concentration of the reactants decreases, there is less particles to form the products therefore less rate.</span>
Answer:
Option B.
Explanation:
As any reaction of combustion, the O₂ is a reactant and the products are CO₂ and H₂O. Combustion reaction for ethane is:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
So 2 moles of ethane react with 7 moles of oxygen to make 4 moles of dioxide and 6 moles of water.
Then 2 moles of ethane will produce 4 moles of CO₂
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
Hello,
Here is your answer:
The proper answer to this question is option D "<span>sodium hydroxide".
Here is how:
Sodium Hydroxide its a white substance that is a </span><span>electrolyte.
Your answer is D.
If you need anymore help feel free to ask me!
Hope this helps!</span>
<u>Answer:</u> The half life of the reaction is 1190.7 seconds
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:
where,
k = rate constant of the reaction = 
= half life of the reaction = ?
Putting values in above equation, we get:
Hence, the half life of the reaction is 1190.7 seconds
