Hydrocarbons are a type of organic compound. Hydrocarbons contains a carbon, hydrogen and oxygen bond that makes it
organic. Chlorides, carbonates and nitrous oxides are not organic compounds but
they are inorganic compounds.
Answer:
35750.4 Joules
Explanation:
Using the formula as follows;
Q = m × c × ∆T
Where;
Q = amount of heat (joules)
m = mass of substance (g)
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C)
According to the provided information,
mass (m) = 320.0 grams
c = 4.2 J/g°C
∆T = (50.8°C - 24.2°C) = 26.6°C
Q = ?
Using; Q = m × c × ∆T
Q = 320 × 4.2 × 26.6
Q = 35750.4 J
Answer: 
for the reaction is -186.75 J/K
Explanation:
Change in entropy () for the given reaction under standard condition is given by-
= ![[3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]](https://tex.z-dn.net/?f=%5B3%5Ctimes%20S_%7Brhombic%7D%5E%7B0%7D_%7B%28s%29%7D%5D%2B%5B2%5Ctimes%20S_%7BH_%7B2%7DO%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B2%5Ctimes%20S_%7BH_%7B2%7DS%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B1%5Ctimes%20S_%7BSO_%7B2%7D%7D%5E%7B0%7D_%7B%28g%29%7D%5D)
So = ![[3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol]](https://tex.z-dn.net/?f=%5B3%5Ctimes%2031.8%20J%2FK.mol%5D%2B%5B2%5Ctimes%20188.825%20J%2FK.mol%5D-%5B2%5Ctimes%20205.79%20J%2FK.mol%5D-%5B1%5Ctimes%20248.22%20J%2FK.mol%5D)
= -186.75 J/K
Answer:
frequency = 0.47×10⁴ Hz
Explanation:
Given data:
Wavelength of wave = 6.4× 10⁴ m
Frequency of wave = ?
Solution:
Formula:
Speed of wave = wavelength × frequency
Speed of wave = 3 × 10⁸ m/s
Now we will put the values in formula.
3 × 10⁸ m/s = 6.4× 10⁴ m × frequency
frequency = 3 × 10⁸ m/s / 6.4× 10⁴ m
frequency = 0.47×10⁴ /s
s⁻¹ = Hz
frequency = 0.47×10⁴ Hz
Thus the wave with wavelength of 6.4× 10⁴ m have 0.47×10⁴ Hz frequency.
Answer:
A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:
C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol
The value given for ΔH∘rxnΔHrxn∘ means that:
a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.
b. the reaction is endothermic.
c. the enthalpy of formation of propane is 2202 kJ/mol.
d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.
e. None of these.
