discuss how variations in electronegativity result in the unequal sharing of electrons in polar molecules.
1 answer:
You might be interested in
A curve of temperature vs. time for the entire heating process.
The sample is heated up to 100.°C, therefore, the heat and time required to heat the sample to its boiling point, the heat and time required to boil the sample, and the heat and time required to heat the sample from its boiling point to 100.°C are needs to be calculated.
i ) Calculating the heat and time required to heat the sample to its boiling point:
Boiling point = 85°C
C(liquid) = 2.5 J/g °C
The heat required up to melting the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from -20°C to 85°C can be calculated as,
Therefore, T f = 85°C and T i = - 20°C
Plug in the values in the specific heat formula to calculate the heat energy required to heat the sample to its melting point,
q3 = 25 g × 2.5 J/g °C × [85 - (-20)]°C
= 25 J/°C ×[85+20]°C
= 6562.5 J
The total heat energy required for heating the sample from initial temperature to boiling point is:-
q1 + q2 + q3 = 500 J + 4500 J + 6562.5 J
= 11562.5 J
The Rate of heating = 450 J/min
450. J = 1 min
11562.5 J = ? min
11562.5 J × 1min/450 J = 25.69 min
ii) Calculating the heat and time required to boil the sample:
∆H Vap = 500 J/g
The boiling is the phase change from liquid to gas at 85°C, therefore, the heat required to boil the sample can be determined
q4= m × ∆Hvap
= 25 g × 500 J/g
= 12500 J
Thus, total heat required to this phase change is q1 + q2 + q3 + q4 = 500 J + 4500 J +6562.5 J + 12500 J = 24062.5 J
The Rate of heating = 450 J / min
450 J = 1 min
24062.5 J = ? min
24062.5J × 1min / 450 J = 53.47 min
iii) Calculating the heat and time required to heat the sample from its boiling point to 100°C
C gas = 0.5 J / g °C
The heat required to boil the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from 85°C to 100°C can be calculated as,
Therefore, T f = 100.°C and T i = 85°C
q5 = 25 g × 0.5 J / g °C × [100 - 85] °C
= 25 J / °C ×15 °C
= 187.5 J
The total heat energy required for heating the sample from initial temperature to 100°C is
q1 + q2 + q3 + q4 + q5 = 500 J + 4500 J + 2625J + 12500 J + 187.5 J
=24250 J
The Rate of heating = 450 J / min
450. J = 1 min
24250 J=? min
Thus, heating the sample to 100.°C takes a total of 53.89 min.
iv) Draw a curve of temperature vs. time for the entire heating process:-
Temperature °C Temperature K Heat energy (J) Time (min)
-40 °C 233 0 0
-20 °C 253 500 1.11
Melting -20 °C 253 5000 11.11
85 °C 358 11562.5 25.69
Boiling 85 °C 358 24062.5 53.475
100 °C 373 24250 53.89
Hence, the graph for the result is in the image.
Learn more about temperature here:-brainly.com/question/24746268
#SPJ4
Answer:
1.65 L
Explanation:
The equation for the reaction is given as:
A + B ⇄ C
where;
numbers of moles = 0.386 mol C (g)
Volume = 7.29 L
Molar concentration of C =
= 0.053 M
A + B ⇄ C
Initial 0 0 0.530
Change +x +x - x
Equilibrium x x (0.0530 - x)
where
K is given as ; 78.2 atm-1.
So, we have:
Using quadratic formula;
where; a = 78.2 ; b = 1 ; c= - 0.0530
= or
= or
= 0.0204 or -0.0332
Going by the positive value; we have:
x = 0.0204
[A] = 0.0204
[B] = 0.0204
[C] = 0.0530 - x
= 0.0530 - 0.0204
= 0.0326
Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326
= 0.0734
Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT
if we make V the subject of the formula; we have:
where;
P (pressure) = 1 atm
n (number of moles) = 0.0734 mole
R (rate constant) = 0.0821 L-atm/mol-K
T = 273.15 K (fixed constant temperature )
V (volume) = ???
V = 1.64604
V ≅ 1.65 L