It turns unto h2o and then back into water
Answer:
The reaction will be slower because the leaving group will be poorer.
Explanation:
One Important thing to put at the back of our mind is that weak bases are good leaving groups. Another thing to take note is that in the halogen series(which is our main subject in this question) as you go down the group the greater or the more heavy the halide is and the heavier halides are more stable that is Bromine will be more stable than chlorine.
Now, we are now told that the bromo halide is been replaced by the chloro halide which means that the rate of Reaction will surely decrease because the leaving group. The cl^- is a stronger base.
Answer: 12.0 milliliters of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn.
Explanation:
moles =
moles of zinc =
The balanced chemical equation is :
According to stoichiometry:
1 mole of zinc reacts with = 2 moles of HCl
Thus 0.0390 moles of zinc reacts with = 
moles of HCl
To calculate the volume for given molarity, we use the equation:
.....(1)
Molarity of 
solution = 6.50 M
Volume of solution = ?
Putting values in equation 1, we get:
Thus 12.0 ml of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn
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
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl
Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.
To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH
Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH
Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH
Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH
I hope this helps.
