Answer:
The answer to your question is: The first option is correct.
Explanation:
From the data given, we conclude that as the amount of salt increases, the boiling point of water increases.
Options
If salt is added to water, the water will boil at a higher temperature This is the hypothesis for the experiment.
Salt makes water boil. This statement is not true.
If water is boiling, it must have salt added to it. This statement is incorrect, it is not the goal of the experiment.
If salt is added to water, the water will get cloudy There is not evidence of that from the information given.
Answer:
fluorine
Explanation:
The element fluorine has 9 protons and 9 electrons.
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.
Formic acid is the simplest carboxylic acid with a structure of HCOOH and has a pka of 3.75. The pka refers to the acidity of the molecule, which in this example refers to the molecules ability to give up the proton of the O-H. A decrease in the pka value corresponds to an increase in acidity, or an increase in the ability to give up a proton. When an acid gives up a proton, the remaining anionic species (in this case HCOO-) is called the conjugate base, and an increase in the stability of the conjugate base corresponds to an increase in acidity.
The pka of a carboxylic can be affected greatly by the presence of various functional groups within its structure. An example of an inductive effect changing the pka can be shown with trichloroacetic acid, Cl3CCOOH. This molecule has a pka of 0.7. The decrease in pka relative to formic acid is due to the presence of the Cl3C- group, and more specifically the presence of the chlorine atoms. The electronegative chlorine atoms are able to withdraw the electron density away from the oxygen atoms and towards themselves, thus helping to stabilize the negative charge and stabilize the conjugate base. This results in an increase in acidity and decrease in pka.
The same Cl3CCOOH example can be used to explain how dipoles can effect the acidity of carboxylic acids. Compared to standard acetic acid, H3CCOOH with a pka of 4.76, trichloroacetic acid is much more acidic. The difference between these structures is the presence of C-Cl bonds in place of C-H bonds. A C-Cl bond is much more polar than a C-H bond, due the large electronegativity of the chlorine atom. This results in a carbon with a partial positive charge and a chlorine with a partial negative charge. In the conjugate base of the acid, where the molecule has a negative charge localized on the oxygen atoms, the dipole moment of the C-Cl bond is oriented such that the partial positive charge is on the carbon that is adjacent to the oxygen atoms containing the negative charge. Therefore, the electrostatic attraction between the positive end of the C-Cl dipole and the negative charge of the anionic oxygen helps to stabilize the entire species. This level of stabilization is not present in acetic acid where there are C-H bonds instead of C-Cl bonds since the C-H bonds do not have a large dipole moment.
To understand how resonance can affect the pka of a species, we can simply compare the pka of a simple alcohol such as methanol, CH3OH, and formic acid, HCOOH. The pka of methanol is 16, suggesting that is is a very weak acid. Once methanol gives up that proton to become the conjugate base CH3O-, the charge cannot be stabilized in any way and is simply localized on the oxygen atom. However, with a carboxylic acid, the conjugate base, HCOO-, can stabilize the negative charge. The lone pair electrons containing the charge on the oxygen atom are able to migrate to the other oxygen atom of the carboxylic acid. The negative charge can now be shared between the two electronegative oxygen atoms, thus stabilizing the charge and decreasing the pka.
