Explanation:
The given data is as follows.
Molecular weight of azulene = 128 g/mol
Hence, calculate the number of moles as follows.
No. of moles = 
= 
= 0.0030625 mol of azulene
Also, 
Now, putting the given values as follows.
= 11748.67 J
So, 
= 1886.4 J
Therefore, heat of reaction will be calculated as follows.
= (11748.67 + 1886.4) J
= 13635.07 J
As,
13635.07 J = 
dE = 
= 4452267.75 J/mol
or, = 4452.26 kJ/mol (as 1 kJ = 1000 J)
Thus, we can conclude that 
for the given combustion reaction per mole of azulene burned is 4452.26 kJ/mol.
Answer:
Find the domain and the range of the following:
x y
3 2
5 7
1 4
9 2
3 7
Explanation:
✅Show work regardless if student got answer correct or incorrect
q = mC∆T
q = (30.0g)(0.900J/goC)(50oC)
q = 1350 J
So, the right answer is 1350 J
IamSugarBee◽
Answer:
1223.38 mmHg
Explanation:
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 
Also,
Moles = mass (m) / Molar mass (M)
Density (d) = Mass (m) / Volume (V)
So, the ideal gas equation can be written as:
Given that:-
d = 1.80 g/L
Temperature = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (32 + 273.15) K = 305.15 K
Molar mass of nitrogen gas = 28 g/mol
Applying the equation as:
P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K
⇒P = 1223.38 mmHg
<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>
Answer:the answer is b
Explanation:I took the test and got it right
