Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
<span>specific heat solns = specific heat H2O = 4.18 J/K*C </span>
<span>CuSO4 + 2KOH = Cu(OH)2 + 2H2O </span>
<span>q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T </span>
<span>q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) </span>
<span>dHrxn in J/mol= q/0.225 mol CuSO4 </span>
<span>Then convert to kJ/mol
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Answer:
Explanation:
Hello!
In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:
Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:
By plugging in the moles and molarity, we obtain:
Which in mL is:
Best regards!
Answer:
Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao
Explanation:
[H+][OH-] = 1x10-14 = Kw (Memorize this relationship)
(2.70x10-2)[OH-] = 1x10-14
[OH-] = 1x10-14/2.70x10-2
[OH-] = 3.70x10-13 M (assuming you ignore the autoionization of H2O)
