<span>Answer:
Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass.
râšM = constant
Therefore for two gases the ratio rates is given by:
r1 / r2 = âš(M2 / M1)
For Cl2 and F2:
r(Cl2) / r(F2) = âš{(37.9968)/(70.906)}
= 0.732 (to 3.s.f.)</span>
Answer:
Explanation:
Hello!
In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:
Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:
Best regards!
The answer I believe is 3.340kj.
Answer: -
Concentration of PbI₂ = 1.5 x 10⁻³ M
PbI₂ dissociates in water as
PbI₂ ⇄ Pb²⁺ + 2 I⁻
So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.
Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =
= 1.5 x 10⁻³ x 2 M
= 3 x 10⁻³ M
PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.
[Pb²⁺] = 1.5 x 10⁻³ M
So solubility product for PbI₂
Ksp = [Pb²⁺] x [ I⁻]²
=1.5 x 10⁻³ x (3 x 10⁻³)²
= 4.5 x 10⁻⁹
<span>are
in random, constant, straight-line motion</span>
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<span>your welcome :)</span>
<span>all I Ask is for a thank u peaceeee</span>
