Answer:
the work input is depented on the work output
Explanation:
The answer
<span>the molar ratio for the following equation
____C3H8 + ____O2 Imported Asset ____CO2 + ____ H2O
</span>after it has been properly balanced:
__1_C3H8 + ____5O2 Imported Asset ____3CO2 + ____ 4H2O
proof:
number of C =3 (C3H8; 3CO2)
number of H =8 (C3H8 ; 4H2O)
number of O = 10(5x2) or (2x3+4) (5O2;4H2O)
the answer is
<span>Reactants: C3H8 = 1, O2 = 8; Products: CO2 = 3 and H2O = 4</span>
Reaction of option c produces precipitate.
Rhodium on reacting with potassium phosphate produces rhodium phosphate which remain in solution due to low lattice energy for rhodium phosphate.
Niobium on reacting with lithium carbonate produces niobium carbonate and it will remain in aqueous form.
Cobalt on reacting with zinc nitrate produces cobalt nitrate. This, Co(NO3 )2 is insoluble precipitate and settles at bottom whereas zinc ion will remain in solution as follows:
Potassium ion on reacting with sodium sulfide produces potassium sulfide which remain in solution
The uranium within these items is radioactive and should be treated with care. Uranium's most stable isotope, uranium-238, has a half-life of about 4,468,000,000 years. It decays into thorium-234 through alpha decay or decays through spontaneous fission.
The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
Read more on equilibrium constant;
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