Answer:
check which reactant is totally consumed and which one remains in the mixture
Explanation:
<em>Apart from doing calculations during an experiment, one can determine which reactant is limiting and which one is in excess by checking the resulting mixture for the presence of reactants.</em>
A limiting reactant is one that determines the amount of product formed during a reaction. It is usually a reactant that is lower than stoichiometry amount.
On the other hand, an excess reactant is one that is present in more than the stoichiometrically required amount during a reaction.
Limiting reactants will be totally consumed in a reaction while excess reactant would still be seen present in mixture after the reaction has stopped.
<u>Hence, apart from using stoichiometric calculation to determine which reactant is limiting or in excess during an experiment, one can just check the final mixture of the reaction for the presence of any of the reactants. The reactant that is detected is the excess reactant while the one without traces in the final mixture is the limiting reactant.</u>
Metal ore has other elements in it as well. Also sediment and stone might cover the ore. We don't want to have a phone with sediment on it do we? thus these few reasons are why.
Answer:
[C] carbon solid
Explanation:
Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.
Answer:
1. 0.125 mole
2. 42.5 g
3. 0.61 mole
Explanation:
1. Determination of the number of mole of NaOH.
Mass of NaOH = 5 g
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
Mole of NaOH =?
Mole = mass /molar mass
Mole of NaOH = 5/40
Mole NaOH = 0.125 mole
2. Determination of the mass of NH₃.
Mole of NH₃ = 2.5 moles
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 2.5 × 17
Mass of NH₃ = 42.5 g
3. Determination of the number of mole of Ca(NO₃)₂.
Mass of Ca(NO₃)₂ = 100 g
Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (3×16)]
= 40 + 2[14 + 48]
= 40 + 2[62]
= 40 + 124
= 164 g/mol
Mole of Ca(NO₃)₂ =?
Mole = mass /molar mass
Mole of Ca(NO₃)₂ = 100 / 164
Mole of Ca(NO₃)₂ = 0.61 mole
Answer:
a. Work, ΔE is negative;
b. Work, ΔE is negative;
c. Work, ΔE is positive.
Explanation:
In the three cases, there is energy exchange in primarily work. The heat is the energy flow because of the difference in temperature. Of course, some heat may be lost in the cases by dissipation.
In the letter <em>a</em> the system is at an initial velocity different from 0, and then it stops. The energy that is represented here is the kinetic energy, which is the energy of the movement. Note that the system goes from a higher velocity to 0, so it is losing kinetic energy, or work, so ΔE = Efinal - Einitial < 0.
In letter <em>b</em>, the system is falling from a certain high to the floor, so its gravitational potential energy is change. That potential energy represents the energy that gravity does when an object shifts vertically. Because it goes from a high to 0, the energy is been lost, so ΔE = Efinal - Einitial < 0.
In letter <em>c</em>, the system is going higher and with higher velocity, so there is a greatness in the gravitational potential energy and the kinetic energy, both works, so ΔE = Efinal - Einitial > 0.
