New answers tagged combustion
2
The excess of $\ce{O2}$ is ${7.32~ mol}$. So the excess air is : ${7.32 ~mol· (1 + 3.76) = 34.84~ mol}$.
The total amount of air is : ${a·4.76 = 19.53~ mol ·4.76 = 92.96~ mol}$.
The percentage of excess air is : $34.84/92.96 = 0.3748 = 37.48$%
0
It really depends exactly on what you're mixing the water with. Soaking a rag or paper towel will reduce the flammability and is recommended by many safety organizations online (here is one). However, if you are washing off a large amount of solvents or flammables with water and the chemicals are immiscible in water, it can sit on the top of the water and ...
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