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
Rubidium is an element that belongs to Group 1. As such it will have physical properties similar to the other Group 1 elements. Rubidium is below
Potassium in the periodic table but above
Cesium. As such it would be most like one of those two elements.
The sample of argon gas that has the same number of atoms as a 100 milliliter sample of helium gas at 1.0 atm and 300 is 100. mL at 1.0 atm and 300. K
The correct option is D.
<h3>What is the number of moles of gases in the given samples?</h3>
The number of moles of gases in each of the given samples of gas is found below using the ideal gas equation.
The ideal gas equation is: PV/RT = n
where;
- P is pressure
- V is volume
- n is number of moles of gas
- T is temperature of gas
- R is molar gas constant = 0.082 atm.L/mol/K
Moles of gas in the given helium gas sample:
P = 1.0 atm, V = 100 mL or 0.1 L, T = 300 K
n = 1 * 0.1 / 0.082 * 300
n = 0.00406 moles
For the argon gas sample:
A. n = 1 * 0.05 / 0.082 * 300
n = 0.00203 moles
B. n = 0.5 * 0.05 / 0.082 * 300
n = 0.00102 moles
C. n = 0.5 * 0.1 / 0.082 * 300
n = 0.00203 moles
D. n = 1 * 0.1 / 0.082 * 300
n = 0.00406 moles
Learn more about ideal gas equation at: brainly.com/question/24236411
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Only the Fe is unbalanced.
:)
