Answer:
the solubility of CaCO3 is 0.015g/l 25 °C
is favored at equilibrium
Explanation:
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?
solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)
CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)
in partial dissociation , we can say
2.25x 10^-8=
let Ca^2+=CO3^-2=S
2.25x10^-8=S*S
S^2=2.25x10^-8
S=0.00015mol/L
Converting that to g/l
the relative molecular mass of CaCO3=100g/mol
0.00015*100g/mol
0.015g/l
the solubility of CaCO3 is 0.015g/l @room temperature
is favored at equilibrium
7, i'm pretty sure, hope I helped!
Molar mass CH4 = 16.0 g/mol
* number of moles:
932.3 / 16 => 58.26875 moles
T = 136.2 K
V = 0.560 L
P = ?
R = 0.082
Use the clapeyron equation:
P x V = n x R x T
P x 0.560 = 58.26875 x 0.082 x 136.2
P x 0.560 = 650.76
P = 650.76 / 0.560
P = 1162.07 atm
<u>2.37</u> is the pH of 0.075 M HZ.
<h3>What is pH?</h3>
The term pH, which originally stood for "potential of hydrogen" (or "power of hydrogen"), is used in chemistry to describe how acidic or basic an aqueous solution is. Lower pH values are summarized for acidic solutions (solutions with higher H+ ion concentrations) than for basic or alkaline solutions.
The pH scale is inversely indicates to the concentration of hydrogen ions in the solution and is logarithmic.
⇒pH = -log(
)
Acidic solutions are those with a pH below 7, and basic solutions are those with a pH above 7, at a temperature of 25 °C (77 °F). At this temperature, solutions with a pH of 7 are neutral (e.g. pure water). The pH neutrality relies on temperature, falling below 7 if the temperature rises above 25 °C.
Learn more about pH
brainly.com/question/12609985
#SPJ4
All mixtures contain only one kind of atom.
~Mschmindy