Answer:
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
Explanation:
Moles of NaOH = ![\frac{15.0 g}{40 g/mol}=0.375 mol](https://tex.z-dn.net/?f=%5Cfrac%7B15.0%20g%7D%7B40%20g%2Fmol%7D%3D0.375%20mol)
Molarity of the nitric acid solution = 0.250 M
Volume of the nitric solution = 0.150 L
Moles of nitric acid = n
![Molarity=\frac{Moles}{Volume(L)}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7BMoles%7D%7BVolume%28L%29%7D)
![n=0.250 M\times 0.150 L=0.0375 mol](https://tex.z-dn.net/?f=n%3D0.250%20M%5Ctimes%200.150%20L%3D0.0375%20mol)
![NaOH+HNO_3\rightarrow NaNO_3+H_2O](https://tex.z-dn.net/?f=NaOH%2BHNO_3%5Crightarrow%20NaNO_3%2BH_2O)
According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :
of NaOH
Moles of NaOH left unreacted in the solution =
= 0.375 mol - 0.0375 mol = 0.3375 mol
![NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)](https://tex.z-dn.net/?f=NaOH%28aq%29%5Crightarrow%20Na%5E%2B%28aq%29%2BOH%5E-%28aq%29)
1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.
Then 0.3375 moles of NaOH will give :
of hydroxide ion
The molarity of hydroxide ion in solution ;
![=\frac{0.3375 mol}{0.150 L}=2.25 M](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.3375%20mol%7D%7B0.150%20L%7D%3D2.25%20M)
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
the answer is 0.000097 KM
Answer:
k = 0.0306 min-1
Explanation:
The table is given as;
Time, Concentration
0 1.48
5 1.27
10 0.98
15 0.84
The integrated rate law for a first order reaction is given as;
ln [A] = -kt + ln [Ao]
where;
[A] = Final Concentration
[Ao] = Initial Concentration
k = rate constant
t = time
In the table, taking the first two sets of values;
t = 5
k = ?
[Ao] = 1.48
[A] = 1.27
Inserting into the equation;
ln(1.27) = - k (5) + ln(1.48)
ln(1.27) - ln(1.48) = -5k
-0.1530 = -5k
k = -0.1530 / -5
k = 0.0306 min-1
It is a heterogeneous mixture <span>
</span>
Only the amount of gas is held constant.