The empirical formula : C₁₂H₄F₇
The molecular formula : C₂₄H₈F₁₄
<h3>Further explanation</h3>
mol C (MW=12 g/mol)
mol H(MW=1 g/mol) :
mol F(MW=19 g/mol)
mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7
Empirical formula : C₁₂H₄F₇
(Empirical formula)n=molecular formula
( C₁₂H₄F₇)n=562 g/mol
(12.12+4.1+7.19)n=562
(281)n=562⇒ n =2
Molecular formula : C₂₄H₈F₁₄
<h3>
Answer:</h3>
322.7 kW
<h3>
Explanation:</h3>
- Power refers to the rate at which work is done.
- Therefore; Power = Work done ÷ time
- It is measured in joules per seconds or Watts
In this case, we are required to convert 0.3227 MW to kilowatts
We need to know that;
- 10^6 watts = 1 Megawatts(MW)
- 10^3 Watts = 1 kilowatts (kW)
Therefore;
10^3 kW = 1 MW
Therefore, the suitable conversion factor is 10^3kW/MW
Hence;
0.3227 MW is equivalent to;
= 0.3227 MW × 10^3kW/MW
= 322.7 kW
Thus, the peak power output is 322.7 kW
The heat released by cpmolete the combustion of organic products with oxygen is called heat of combustion.
Here 1 mol of CH4 realesed 802.3 KJ
To emit 264 kJ you multiply you need (1 mol of CH4/802.3 kJ)* 264 kJ = 0.329 mol of CH4
The molar mass, MM, of CH4 is 12 g/mol + 4*1g/mol = 16 g/mol
The to obtain the mass multiply the number of moles times the molar mass:
mass = n * MM = 0.329mol * 16g/mol = 5.26 grams
Answer: 5.26 grams
