Explanation:
a. Adding a catalyst
no effect
.( Catalyst can only change the activation energy but not the free energy).
b. increasing [C] and [D]
Increase the free energy
.
c. Coupling with ATP hydrolysis
decrease the free energy value
.
d.Increasing [A] and [B]
decrease the free energy.
Answer:
13000
Explanation:
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
A gram is a unit of weight equal to 1/1000th of a kilogram. A gram is the approximate weight of a cubic centimeter of water.
Answer:
In this phenomenon we talk about ideal gases, that is why in these equations the constant is the number of moles and the constant R, which has a value of 0.082
Explanation:
The complete equation would have to be P x V = n x R x T
where n is the number of moles, and if it is not clarified it is because they remain constant, as the question was worded.
On the other hand, the symbol R refers to the ideal gas constant, which declares that a gas behaves like an ideal gas during the reaction, and its value will always be the same, which is why it is called a constant. The value of R = 0.082.
The ideal gas model assumes that the volume of the molecule is zero and the particles do not interact with each other. Most real gases approach this constant within two significant figures, under pressure and temperature conditions sufficiently far from the liquefaction or sublimation point. The real gas equations of state are, in many cases, corrections to the previous one.
The universal constant of ideal gases is not a fundamental constant (therefore, choosing the temperature scale appropriately and using the number of particles, we can have R = 1, although this system of units is not very practical)
Answer:
(A) 15.0 °C
Explanation:
The water in beaker A gains heat because its initial temperature (10 °C) is less than the initial temperature of the water in beaker B (20 °C) which loses heat.
Let T3 be the final temperature
Heat gained by beaker A = heat loss by beaker B
mc(T3 - T1) = mc(T2 - T3)
The mass and specific heat of water in both beakers are the same. Therefore, (T3 - T1) = (T2 - T3)
T1 is initial temperature of beaker A = 10 °C
T2 is initial temperature of beaker B = 20 °C
T3 - 10 = 20 - T3
T3 + T3 = 20 + 10
2T3 = 30
T3 = 30/2 = 15 °C
