Answer:Option (3) less than the sum of its components' masses
Explanation:
Answer:
okay so lets say i drop a ball that weighs 2 pounds off a building and a ball that ways 1 pound off the building at the same time the lighter ball would hit the ground first because its lighter and so they heavier it is the slower it drops the lighter it is the faster it drops
Explanation:
The given question is incorrect. The correct question is as follows.
If 20.0 g of
and 4.4 g of
are placed in a 5.00 L container at
, what is the pressure of this mixture of gases?
Explanation:
As we know that number of moles equal to the mass of substance divided by its molar mass.
Mathematically, No. of moles = 
Hence, we will calculate the moles of oxygen as follows.
No. of moles = 
Moles of
=
= 0.625 moles
Now, moles of 
= 0.1 moles
Therefore, total number of moles present are as follows.
Total moles = moles of
+ moles of 
= 0.625 + 0.1
= 0.725 moles
And, total temperature will be:
T = (21 + 273) K = 294 K
According to ideal gas equation,
PV = nRT
Now, putting the given values into the above formula as follows.
P =
= 
=
atm
= 3.498 atm
or, = 3.50 atm (approx)
Therefore, we can conclude that the pressure of this mixture of gases is 3.50 atm.
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
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