Answer:
Angle ABE measures 27°.
Explanation:
Refer to the attached diagram related to this question.
The given values are ∠ABE=2n+7 and ∠EBF=4n-13.
Clearly seen in the diagram, ∠ABE and ∠EBF are equal in measure.


Move variable components to one side of the equation.


Split both sides by 2.

The solution for n arrives at 10.
The next step is to calculate ∠ABE.

Consequently, the measurement of angle ABE is 27°.
Given:
Mass of the ionic compound = 10.00 g
Mass of water = 75.0 g
Initial temperature of water T1= 23.2 C
Final temperature of water T2 = 31.8 C
Specific heat of water c = 4.18 J/gC
To determine:
Enthalpy of dissolution of the ionic compound
Heat gained by water equation:
Q = mcΔT
m = mass of water
c = specific heat
ΔT = change in temperature (T2-T1)
Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J
Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution)
Therefore, q(ionic) = 2696 J
ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g
Answer: A) enthalpy change = 2.7*10² J/g
Fe 3+ + SCN- --> FeSCN 2+
<span>.......Fe 3+.......SCN-.........FeSCN 2+ </span>
<span>I.......0.04..........0.001.............. </span>
<span>C........-x...............-x............. </span>
<span>E.....0.04-x.....0.001-x...........x </span>
<span>Keq = 203.4 = x / (0.04-x)(0.001-x) </span>
<span>203.4 = x / (x^2 - 0.041x + 4x10^-5) </span>
<span>203.4x^2 - 8.34x + 0.00094 = x </span>
<span>203.4x^2 - 9.34x + 0.00094 = 0 </span>
<span>x = -0.0001M or 0.0458M </span>
<span>therefore, according to the calculated Keq, all of the SCN- and Fe 3+ would be fully converted into FeSCN 2+</span>