CH₇ is the empirical formula of the car fuel.
Explanation:
To find the empirical formula we use the following algorithm.
First divide each mass the the molar weight of each element:
for carbon 2.87 / 12 = 0.239
for hydrogen 3.41 / 2 = 1.705
And now divide each quantity by the lowest number which is 0.239:
for carbon 0.239 / 0.239 = 1
for hydrogen 1.705 / 0.239 = 7.13 ≈ 7
The empirical formula of the car fuel is CH₇.
I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.
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Answer:
i think that the answer is solid liquid and gas.
Explanation:
Answer:
2Ca + O₂ → 2CaO
Explanation:
Cuando el Calcio (Ca) reacciona con oxígeno (O₂) se produce cal (CaO). La cal es un sólido inodoro de color blanco a grisáceo. La cal es un óxido que se encuentra presente en el cemento y su cuantificación permite determinar la calidad y el tipo de cemento a utilizar.
La reacción que describe el proceso anterior es:
Ca + O₂ → CaO
Para balancear los oxígenos, se deben poner 2 CaO como producto:
Ca + O₂ → 2CaO
Para balancear los calcios, se pone como coeficiente del Ca un 2:
<h3>2Ca + O₂ → 2CaO</h3><h3 />
Esta última es la reacción que describe el proceso anterior
Answer: 0.4 g
Explanation:
1) Balanced chemical equation:
2) 2C2H2 + 5 O2 → 4CO2 + 2 H2O
3) mole ratios:
2 mol C2h2 : 5 mol O2
4) Convert 0.13 g C2H2 into number of moles
n = mass in grams / molar mass
molar mass C2H2 = 2 *12g/mol + 2*1 g/mol = 26 g/mol
n = 0.13 g / 26 g/mol = 0.005 mol
5) Set the proportion with the unknown
5 mol O2 x
----------------- = ---------------
2 mol C2H2 0.005 mol C2H2
x = 0.005 mol C2H2 * 5 mol O2 / 2 mol C2H2 = 0.0125 mol O2
6) Convert 0.0125 mol O2 to grams
mass = number of moles * molar mass
molar mass of O2 = 32 g/mol
mass = 0.0125 mol * 32 g/mol = 0.4 g
Answer: 0.4 g
