The plants give out oxygen so for us humans will live
The number of bacteria is given by:
N(t) = N(o) x 2ⁿ
Where N(t) is the number after n hours have passed and N(o) is the original number which is 15.
The number grown in the 12th hour is the difference in the number after the 11th and the 12th hour. Thus:
15 x 2¹² - 15 x 2¹¹
= 30,720 bacteria
Answer: The concentration of hydrogen ions for this solution is
.
Explanation:
Given: pOH = 11.30
The relation between pH and pOH is as follows.
pH + pOH = 14
pH + 11.30 = 14
pH = 14 - 11.30
= 2.7
Also, pH is the negative logarithm of concentration of hydrogen ions.
![pH = - log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D)
Substitute the values into above formula as follows.
![pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5C2.7%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20of%20H%5E%7B%2B%7D%20%3D%201.99%20%5Ctimes%2010%5E%7B-3%7D)
Thus, we can conclude that the concentration of hydrogen ions for this solution is
.
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.


There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6