Answer:
Energy in the campfire originates from the potential chemical energy of the wood, before it is burnt to warm and give light around the campfire.
Explanation:
For a camp fire, the energy input is in the form of the potential chemical energy, stored up in the firewood used to fuel the flame.
The energy output is in the form of heat energy that the campfire radiates all around, light energy given off from the flame, and a little bit of sound energy, heard in the cracking of the firewood as they burn in the flame.
chemical energy ⇒ heat energy + light energy + sound energy
You only need a 2 at the end In front of the NaCl
<em><u>examples of metalloids :
</u>Boron,Silicon ,Germanium are some metalloids ...
<u>what they do:
</u>they are used to form the semiconductors and these semiconductors are used in modern computer technology like in circuits ,chips and computer based gadgets... :) <u>
</u></em>
We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.
Dividing the mass of each reactant by its molar mass:
(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6
(10 g O2)(31.999 g/mol) = 0.3125 mol O2.
Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.
Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).
So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.
Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.
