Answer:
333.7 g.
Explanation:
- The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>
Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).
Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>
molaity (m) is the no. of moles of solute per kg of the solvent.
∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))
∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.
∵ no. of moles = mass/molar mass.
<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>
Explanation:
a. Adding a catalyst
no effect
.( Catalyst can only change the activation energy but not the free energy).
b. increasing [C] and [D]
Increase the free energy
.
c. Coupling with ATP hydrolysis
decrease the free energy value
.
d.Increasing [A] and [B]
decrease the free energy.
<span>Cu⁺ is the only one of the ions in the list that will show 8 electrons in a d sublevel....its configuration will be Ar| 4s² 3d⁸
hope this helps</span>
Answer:
4.86×10^23 molecule of Pb
Explanation:
Based on that equation, for every 2 moles of ammonia, you get 3 moles of lead.
So:
2 mol NH3/ 3 mol Pb
Using this ratio we can find the amounts of either molecule. Given 5.38 mol NH3:
(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb
Then, we just need to use Avagadro's number to get the number of molecules.
(8.07)(6.02×10^23) = 4.86×10^23 molecule of Pb
