Answer:
A.'C
Explanation:
Please answer my question
Answer:
%
Explanation:
The ethanol combustion reaction is:
→
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:
Dividing the previous equation by x:
We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:
Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:
Calculate the number of moles of CO2 and water considering the same:
The total number of moles at the reactor output would be:
So, the oxygen mole fraction would be:
%
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
Use Planck's equation (E=hv) to solve. where <span>frequency (v) of ultrviolet radiation is 6.8 × 1015 1/s. </span><span>
</span>The variable h is a
constant equal to 6.63 × 10-34 J·s
E= <span>(6.8 × 1015 1/s)x(</span>6.63 × 10-34 J·s)
Density = 0.601 g/mL
Volume = 7.25 mL
D = m / V
0.601 = m / 7.25
m = 0.601 x 7.25 => 4.35725 g
hope this helps!