A solution is prepared by adding 1.43 mol of potassium chloride (kcl) to 889 g of water. The concentration of kcl is 1.61 molal.
mol of Kcl (potassium chloride)= 1.43
water = 889 g
the formula for calculating molality is:
molality = moles of solute/kilograms of solvent
1kg = 1000g so, 889g = 0.889kg
m = 1.43/0.889 = 1.61 molal
The answer is evaporation<span>.Evaporation is the vaporization of the liquid from its surface into gaseous phase, without boiling the liquid. When all the liquid has passed to gaseous phase the salt dissolved in the salt water will remain as solid crystals.</span><span />
The equilibrium constant expression for KSP of Sr3(PO4)2 is
KSP={(Sr^2+)^3 (PO4^3-)^2/ Sr3(PO4)2}
Explanation
write the ionic equation for Sr3(PO4)2
Sr3(PO4)2 → 3Sr^2+ + 2 PO4^3-
KSP is given by (concentration of the products raised to their coefficient /concentration of reactants raised to their coefficient)
Answer:
Explanation:
Question 7.
We can use the Combined Gas Laws to solve this question.
a) Data
p₁ = 1.88 atm; p₂ = 2.50 atm
V₁ = 285 mL; V₂ = 435 mL
T₁ = 355 K; T₂ = ?
b) Calculation
Question 8. I
We can use the Ideal Gas Law to solve this question.
pV = nRT
n = m/M
pV = (m/M)RT = mRT/M
a) Data:
p = 4.58 atm
V = 13.0 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 385 K
M = 46.01 g/mol
(b) Calculation
