Answer:
To release 7563 kJ of heat, we need to burn 163.17 grams of propane
Explanation:
<u>Step 1</u>: Data given
C3H8 + 5O2 -----------> 3CO2 + 4H2O ΔH° = –2044 kJ
This means every mole C3H8
Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)
<u>Step 2: </u>Calculate the number of moles to produce 7563 kJ of heat
1 mol = 2044 kJ
x mol = 7563 kJ
x = 7563/2044 = 3.70 moles
To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8
<u>Step 3: </u>Calculate mass of propane
Mass propane = moles * Molar mass
Mass propane = 3.70 moles * 44.1 g/mol
Mass propane = 163.17 grams
To release 7563 kJ of heat, we need to burn 163.17 grams of propane
This is categorized as a combustion reaction.
Wavelength= velocity/frequency
wavelength= (3.0 x 10^8m/s) / 7.5 x 10^12 Hz)
you can do the math
I am assuming u that 108 is 10^8 and the 1012 is 10^12
Answer:
0.11%
Explanation:
Without mincing words, let us dive straight into the solution to the question/problem. The first step to solve this question is to write out the chemical reaction, that is the reaction showing the dissociation of acetic acid.
CH3COOH <=======================================> CH3COO⁻ + H⁺
Initially, the amount present in the acetic acid which is = 12M, the concentration for CH3COO⁻ and H⁺ is 0 respectively.
At equilibrium, the amount present in the acetic acid which is = 12 - x, the concentration for CH3COO⁻ = x and H⁺ = x respectively. Note that the ka for acetic acid = 1.8 × 10⁻⁵.
1.8 × 10⁻⁵ = x²/ 14 - x. Therefore, x = 0.0158 M.
The next thing to do is to calculate for the percentage of dissociation, this can be done as given below:
percentage of dissociation = x/14 × 100. Recall that the value that we got for x = 0.0158 M. Hence, the percentage of dissociation = 0.0158 M/ 14m × 100 = 0.11%
