Answer:
The right statement is "The percent transmittance of light, %T, will be higher for Sample 1 compared to Sample 2."
Explanation:
The Beer-Lambert Law states that the absorbance is directly proportional to the concentration of the substance.
Absorbance = εLc
ε: molar absorptivity coeficient
L: optical path length
c: molar concentration
Therefore, if ε and L are the same, the higher the concentration, the higher the absorbance and the lower the transmitance.
Since sample 1 has a higher concentration of CuSO₄ than sample 2, sample 1 has a higher absorbance and lower percent transmittance.
<span>M(HCl) * </span><span>V(HCl) </span>= <span>M(NaOH) * </span><span>V(<span>NaO<span>H)
</span></span></span>
M(HCl) = 0.35
<span>V(HCl) = 45mL
</span>M(NaOH)= 0.35
now, solne for V(NaOH) by putting these values in the above equation.
M(HCl) * <span>V(HCl) </span>= <span>M(NaOH) * </span><span>V(NaOH)</span>
<span>0.35 * 45 = 0.35 * V(NaOH)</span>
<span>V(NaOH) = 45 mL</span>
Answer:fH = - 3,255.7 kJ/mol
Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol