Here we have to choose the property of the liquid, among the given, which the engineer should primary look to use as the fluid of the engine.
The engineer should primarily looked the liquid which have B. low specific heat capacity.
The Specific heat capacity of a substance is equivalent to the heat needed to increase 1⁰ temperature of the unit mass of the substance. Now if the specific heat capacity of the liquid is low then it will need very little amount of energy to transfer the heat.
A. The low thermal conductivity will cause more time to exchange the energy. Which is not required for the engine.
C. There is no relation between the reactivity of the liquid with the use of the same in the engine.
D. The internal energy will also not affect the rate of the cooling or heating of the liquid.
E. The high density also cannot affect the low rate of exchange energy.
Answer:
No
Explanation:
AgNO₃ + KCl --> AgCl + KNO₃
This reaction is a precipitation reaction. AgCl is formed as a white solid due to its low solubility in water.
The answer is to just literally turn it off
Answer:
B) Melting of a solid.
Explanation:
- Kindly, see the attached image.
- It is known that the substance has different 3 phases (solid, liquid, and gas).
- By heating, the substance converted from solid phase to liquid phase passing by a plateau that represents equilibrium between the two phases.
- By further heating, the solid completely converted into liquid phase, then converted to gaseous phase passing by a plateau that represents an equilibrium between the liquid an gaseous phase.
<em>The plateau at the lower temperature represent: B) Melting of a solid.</em>
Answer:
K = 8.1 x 10⁻³
Explanation:
We are told here that these gas phase reactions are both elementary processes, thus the reactions forward and reverse are both first order:
A→B Rate(forward) = k(forward) x [A]
and for
B→A Rate(reverse) = k(reverse) x [B]
At equilibrium we know the rates of the forward and reverse reaction are equal, so
k(forward) x [A] = k(reverse) x [B] for A(g)⇌B(g)
⇒ k(forward) / k(reverse) = [B] / [A] = K
4.7 x 10⁻³ s⁻1 / 5.8 x 10⁻¹ s⁻¹ = 8.1 x 10⁻³ = K
Notice how this answer is logical : the rate of the reverse reaction is greater than the forward reaction ( a factor of approximately 120 times) , and will be expecting a number for the equilibrium constant, K, smaller than one where the reactant concentration, [A], will prevail.
It is worth to mention that this is only valid for reactions which are single, elementary processes and not true for other equilibria.
