I don't fully understand your question, but I believe that plants have functions that are vital to the planet. They take sunlight, make it into food for themselves, and release oxygen that we need to survive.
the amount of heat produced from the combustion of 24.3 g benzene (c6h6) is ΔH = -976.5 kJ
There are two moles of benzene involved in the process (C6H6). Since the heat of this reaction is -6278 kJ, the burning of 2 moles of benzene will result in a heat loss of 6278 kJ. This reaction is exothermic.
Enthalpy, or the value of H, is a unit of measurement for heat that relies on the amount of matter present (number of moles).
Thus, 24.3 g of benzene contains:
n = mass/molar mass, where n = 24.3/78.11, and n = 0.311 moles.
2 moles = 6278 kJ
0.311 moles =x
By the straightforward direct three rule:
2x = -1953.08 x = -976.5 kJ
Learn more about combustion here-
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Answer: 1. HYDROCARBONS? 2. ALKANES?
i'm not exactly AMAZING at this but i did some research and this is what i think it is i'm also not in this grade but i tried.
Answer;
=259 ml
Explanation;
-According to Gay Lussac's Law of Combining Volumes when gases react, they do so in volumes which have a simple ratio to one another, and to the volume of the product formed if gaseous, provided the temperature and pressure remain constant.
-Thus; from the volume of nitrogen and oxygen gases; we have; 316 / 178 = 1.775 moles of nitrogen gas per mole of oxygen gas.
-Therefore, nitrogen gas is the limiting reactant, and for each mole of nitrogen gas used, we will get 1 mole of N2O. This means the resulting volume of N2O with 100% yield will be the same as the volume of nitrogen gas used, thus, 100% yield will produce 316 mL.
However, with 82% yield the volume would be; 316 × 82/100 =259 ml
Therefore; the volume of N2O at 82% yield will be 259 ml
#1- Identify a problem
#2- Collect info on your problem
#3- Make a hypothesis
#4- Design an experiment to test your hypothesis
#5- Collect data and observations
#6- Accept or reject your hypothesis
#7- Record results
Hope this helps.
