First, you have to find now many moles of octane are present in 191.6g of octane. To do this you need to do this you need to divide 191.6g by its molar mass (which is 114g/mol). This will give you 1.681 moles of octane. Then you need to use the fact that 2 moles of octane are us ed to make 16 moles of carbon dioxide to find how many moles of carbon dioxide 1.681mole of octane produces. To do this you need to multiply 1.681mole by 16/2 to get 13.45mol carbon dioxide. The final step is to find the number of grams presswnt in 13.446 moles of carbon dioxide. To do this you need to multiply 13.446 mole by carbon dioxides molar mass (which is 44g/mol) to get 591.6 g of carbon dioxide.
Therefore, 591.6g of carbon dioxide is produced when 191.6 grams of octane is burned.
I hope this helps. Let me know in the comments if anything is unclear.
There is 3.58 He in the balloon.
translate please I'm confused
Explanation:
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Answer:
H2(g)+I2(s)→2HI(s)
Explanation:
Hello there!
In this case, according to the given information and unbalanced chemical reaction, we infer it must be balanced in agreement with the law of conservation of mass because the reactants side has two hydrogen and iodine atoms whereas the products side has just one. In such a way, by placing a 2 on HI, we obtain the following balanced reaction:
H2(g)+I2(s)→2HI(s)
Regards!
Answer:
I'm not sure but all I can find is this for you :)
The temperature of the oxygen gas is 243.75 K.
Using ideal gas law to explain the answer, the absolute temperature of the gas will decrease if the number of moles of the gas increases and it will increase if the volume and/or pressure of the gas increases.
