Answer:
One can determine the specific heat of the metal through using the clarimeter, water, thermometer and using heat equations.
Explanation:
You can learn about heat effects and calorimetery through a simple experiment by boiling water and heating up the metal in it. Then, pour it into your calorimeter and the heat will flow from the metal to the water. The two equlibria will meet: the metal will loose heat into its surroundings (the water) and teh water will absorb the heat. The heat flow for the water is the same as it is for the metal, the only difference being is the negative sign indicating the loss of the heat of the metal.
In terms of theromdynamics, we can deteremine the heat flow for the metal becasue it would be equal to the mangnitued but opposite in direction. Thus, we can say that the specific heat of water qH2O = -qmetal.
The best answer among the choices is option D. New substances are being formed will best explain that this process is a chemical change. Although, it cannot be seen directly we knew that an acid and base will undergo neutralization reaction forming a salt and water. The heat given off by the reaction can be a parameter however it does not only happens for reactions. It can also happen when dissolving substances.
Answer:
the description of a mineral species usually includes its common physical properties such as habit, hardness, lustre, diaphaneity, colour, streak, tenacity, cleavage, fracture, parting, specific gravity, magnetism, fluorescence, radioactivity, as well as its taste or smell and its reaction to acid.
<span>The escape of gas through a small hole in a container is called effusion. This phenomenon happens when the diameter of the hole is small enough compared to the mean free path of the gas particles. This is governed by Graham's Law which states that the rate of effusion is inversely proportional to the molecular weight of the gas.</span>
Answer:
- What is the AGⓇ of this reaction? 0.
- Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
- What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
- If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.
Explanation:
1. To calculate the delta G of a reaction given the K, we use the following equation:
ΔG°= -RT ln K.
Which gives us 0 when K is 1.
2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.
3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.
4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.
