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sladkih [1.3K]
3 years ago
13

Write the charge balance for a solution that contains h , oh–, mg2 , hso4–, so42–, mg(hso4) , mg(oh) , na , and no3–.

Chemistry
1 answer:
Olenka [21]3 years ago
5 0
A solution is when a solute is added to a water solvent. Since water is neutral, the charge of the said solution is dictated by the charge of the ion. This is the subscript of the element. If it is not stated, then we use the common ion form.

1. H: The charge for hydrogen ion is +1.
2. OH⁻: The charge is ⁻1.
3. Mg2: This ion has a charge of +2.
4. HSO₄⁻: -1
5. SO₄²⁻: -2
6. Mg(HSO₄): This is not an ion, but an undissociated compound. So, it has a neutral charge of 7.
7. Mg(OH): Neutral at pH 7
8. Na: Sodium ion has a charge of +1.
9. NO₃⁻: -1.
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MArishka [77]

Answer:

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7 0
2 years ago
Consider the pka (3.75) of formic acid, h-cooh as a reference. with appropriate examples, show how inductive, dipole, and resona
Luden [163]
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.
3 0
3 years ago
Which of the following would be the best representation of input energy and then
Rom4ik [11]
A is the correct answer
4 0
2 years ago
A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.
likoan [24]

Given :

A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

To Find :

The volume of the gas after it is heated.

Solution :

Since, their is no information about pressure in the question statement let us assume that pressure is constant.

Now, we know by ideal gas equation at constant pressure :

\dfrac{V_1}{V_2} = \dfrac{T_1}{T_2}\\\\\dfrac{3.82}{V_2}= \dfrac{204.9}{304.8}\\\\V_2 = \dfrac{304.8}{204.9} \times 3.82\\\\V_2 = 5.68 \ L

Hence, this is the required solution.

3 0
3 years ago
A sample of a compound contains ​0.300 mol​ ​of carbon​ and ​1.20 mol​ ​of hydrogen​. What is the ​empirical formula of the comp
Margarita [4]

Answer:

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Explanation:

The number of moles of carbon and hydrogen has been given as follows:

C = ​0.300 mol​ ​

H = ​1.20 mol

Next, we divide each mole value by the smallest (0.300)

C = 0.300 ÷ 0.300 = 1

H = 1.20 ÷ 0.300 = 4

The empirical ratio of Carbon and Hydrogen is 1:4, hence, the empirical formula is CH4

3 0
3 years ago
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