<h3>Solution-:</h3>
- option D
- maintains a constant volume.
#<em>o</em><em>f</em><em>f</em><em>i</em><em>c</em><em>a</em><em>i</em><em>l</em><em> </em><em>Nazo</em>
<em>ll </em><em>Radhe</em><em> Radhe</em><em> ll</em>
The solubility equilibrium of
:
[tex] CaCrO_{4}(aq)<===>Ca^{2+}(aq) + CrO_{4}^{2-}(aq)\\
Q_{sp}=[Ca^{2+}][CrO_{4}^{2-}]\\
= (0.0200 M)(0.0300 M) \\
= 0.0006
Ksp (0.00071) > Qsp (0.0006). So, <u>no precipitate would form</u>.
Explanation:
here's the answer to your question
double arrow just mean that it's a reversible process, and the reaction can go back and forth.
First let us compute for the number of moles of butane
(molar mass = 58.12 g/mol)
number of moles = 145 g / (58.12 g/mol) = 2.49 mol
<span>We use the ideal
gas equation to calculate the volume:</span>
<span> V = n R T / P</span>
V = 2.49 mol * 62.36367 L torr / mol K * 308.15 K / 745
torr
<span>V = 64.35 L</span>