Answer:
A. Soaps react with ions in hard water to create a precipitate.
B. Soaps are both hydrophobic and hydrophilic.
D. Soaps should be weakly alkaline in solution.
Explanation:
A. Hard water contains <u>magnesium and calcium minerals</u> like calcium and magnesium carbonates, sulfates and bicarbonates. As soon as these minerals come in contact with soap their ions like Mg²⁺ & Ca²⁺ form precipitates.
B. Soap are both hydrophilic and hydrophobic. They reason why they exhibit both the properties is really important for their functionality. The hydrophobic part of soap makes interaction with oil/dust particles while the hydrophilic part makes interaction with water. When the cloth is rinsed the dirt/soap particles are removed from the dirty clothes thereby making them clean.
C. Soaps have alkaline pH i.e. more than 7 that is why they have bitter taste.
The formula for GPE is PE=mgh, where “m” is the mass of the object, “g” is the acceleration due to gravity (~9.8 m/s^2 on Earth’s surface), and “h” is the height of the object from the ground. Therefore,
PE=mgh
PE=(6 kg)(9.8 m/s^2)(1 m)
PE=58.8 kg•m^2/s^2 or 58.8 Newtons
The GPE of the bowling ball under these conditions would be about 59 Newtons.
Hope this helps!
Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14
The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).
pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:
pOH= - log [OH⁻]
On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.
LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:
[LiOH]= [OH-]= 0.073 M
Replacing in the definition of pOH:
pOH= -log (0.073 M)
<u><em>pOH= 1.14 </em></u>
In summary, the pOH of the aqueous solution is 1.14
Learn more:
Mass percentage of a solution is the amount of solute present in 100 g of the solution.
Given data:
Mass of solute H2SO4 = 571.3 g
Volume of the solution = 1 lit = 1000 ml
Density of solution = 1.329 g/cm3 = 1.329 g/ml
Calculations:
Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g
Therefore we have:
571.3 g of H2SO4 in 1329 g of the solution
Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987
Mass percentage of H2SO4 (%w/w) is 42.99 %
From what is given, double replacement is what it is
