Well what are all the calculations you have
To answer this question, we will use pressure law which states that:
"At constant volume of a fixed mass of gas, the pressure of the gas is directly proportional to the temperature"
This means that:
Pi / Ti = Pf / Tf where:
Pi is the initial pressure = 1 atm
Ti is the initial temperature = 100 + 273 = 373 degree kelvin
Pf is the final pressure that we want to calculate
Tf is the final temperature = 139 + 273 = 412 degree kelvin
Substitute in the equation with the givens to calculate the pressure as follows:
1 / 373 = Pf / 412
Pf = (1/373) * 412 = 1.104557641 atm
Answer:
a) CaF₂.
b) 7.81g of CaF₂ are present.
Explanation:
a) The calcium ion has as charge Ca²⁺ and fluoride ion is F⁻, that means formula unit is:
CaF₂
b) 1 molecule of CaF₂ contains 2 anions, F⁻. Thus, the moles of CaF₂ is:
1.2x10²³ anions * (1 moleculeCaF₂ / 2 anions) = 6x10²² molecules of CaF₂
6.022x10²³ molecules = 1mol:
6x10²² molecules of CaF₂ * (1mol / 6.022x10²³molecules) = 0.10 moles CaF₂.
1 mole of CaF₂ has a mass of 78.07g:
0.10 moles CaF₂ * (78.07g / mol) =
7.81g of CaF₂ are present
Answer:
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Explanation:
hope this helps
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
