The precipitation reaction of sodium nitrate and lead (II) chloride would not happen. No reaction will happen between these substances. Hope this answers the question. Have a nice day.
Alkali are soluble bases, however not all bases are soluble in water, therefore not all bases are Alkali.
Supposing a temperature of 25 degrees and supposing that all
activity coefficients are 1
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
Thus a pH of 2.50 would mean that the [H+], the concentration of the hydrogen
ion, would be 10^(-2.50)
pH + pOH = 14
pOH = 14 - pH = 14 - 2.5 = 11.5
MOH- levels would be coordinated with pOH
pOH = -log[OH-] ==> [OH-] = [MOH-] = 10^-pOH = 10^-11.5 = 3.2 x 10^-12
Therefore, MOH¯ = 3.2 × 10¯12 M
Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>
Answer:
2.03
Explanation:
Let's <u>assume we have 1 L of the solution</u>:
- There would be 2.07 ethylene glycol moles.
- The solution would weigh (1000 mL * 1.02 g/mL) = 1020 g.
With that information we can <u>calculate the molality</u>:
- molality = moles of solute / kg of solvent
- molality = 2.07 moles / (1020 ÷ 1000) = 2.03 m
Keep in mind that this is only an estimate, as we used the kg of the solution and not of the solvent.