Answer:
0.02405 g/L is the solubility of argon in water at 25 °C.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
where,
= Henry's constant = 
= partial pressure of carbonated drink = 0.51atm
Putting values in above equation, we get:
Molar mass of argon = 39.95 g/mol
Solubility of the argon gas :
0.02405 g/L is the solubility of argon in water at 25 °C.
D = M/V = 76g / 22ml = 3.4g/ml
Half ~ D = 38g / 11ml = 3.4g/ml
Even if the object you had was cut in half, it’s density would remain the same.
Answer is: the amu of a sodium atom is 23.
The unified atomic mass unit (amu) is a standard unit of atom mass.
One unified atomic mass unit is approximately the mass of one nucleon (proton or neutron).
The unified atomic mass unit is equal 1.66·10⁻²⁷ kg.
n(p⁺) = 11; number of protons in sodium atom.
n(n°) = 12; number of neutrons of sodium atom.
n(p⁺ + n°) = 11 + 12.
n(p⁺ + n°) = 23; number of nucleons in sodium atom.
Answer is: Ksp = 4s³.
Balanced chemical reaction (dissociation) of strontium hydroxide:
Sr(OH)₂(s) → Sr²⁺(aq) + 2OH⁻(aq).
Ksp(Sr(OH)₂) = [Sr²⁺]·[OH⁻]².<span>
[</span>Sr²⁺] = s.<span>
[</span>OH⁻] = [Sr²⁺] = 2s<span>
Ksp = (2s)² · x = 4s³.
Ksp is the solubility product constant for
a solid substance dissolving in an aqueous solution.
[</span>Sr²⁺]
is equilibrium concentration of iumcations.<span>
[</span>OH⁻] is equilibrium concentration of hydroxide anions.
Answer:
Volume of sample after droping into the ocean=0.0234L
Explanation:
As given in the question that gas is idealso we can use ideal gas equation to solve this;
Assuming that temperature is constant;
Lets 
and 
are the initial gas parameter before dropping into the ocean
and 
and 
are the final gas parameter after dropping into the ocean
according to boyle 's law pressure is inversly proportional to the volume at constant temperature.
hence,
P1=1 atm
V1=1.87L
P2=80atm
V2=?
After putting all values we get;
V2=0.0234L
Volume of sample after droping into the ocean=0.0234L
