Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
Answer:
im pretty sure its 2...
Explanation:
if its wrong im sorry
if its right brainliest pls?
Answer:
energy = heat = producers = consumer animals =
decomposers = heat
Explanation:
this is the best
Answer:
The purpose of the experiment is to see how water of different temperature and salinity affect the density.
Explanation:
Temperature and salinity directly affect the density of the water. Water of low temperature is more dense than water of high temperature, BUT, (fresh)water with no salt is less dense than (sea)water with more salt, so temperature and salinity change density of water.
