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alexandr1967 [171]
3 years ago
9

Many homes that are not

Chemistry
1 answer:
andrew11 [14]3 years ago
6 0

Answer:

pump that is the answer hope its right

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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
Why are alkali metals so reactive?
Savatey [412]

Answer:

<em>Alkali metals are among the most reactive metals. This is due in <u>part to their larger atomic radii and low ionization energies.</u> They tend to donate their electrons in reactions and have an oxidation state of +1. ... All these characteristics can be attributed to these elements' large atomic radii and weak metallic bonding.</em>

Explanation:

<em>I </em><em>hope</em><em> it</em><em> will</em><em> help</em><em> you</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

<em>#</em><em>C</em><em>A</em><em>R</em><em>R</em><em>Y</em><em>O</em><em>N</em><em>L</em><em>E</em><em>R</em><em>A</em><em>N</em><em>I</em><em>N</em><em>G</em>

8 0
2 years ago
How much heat do you need to raise the temperature of 150 g of ice from -30°C to -15°c?
OLga [1]
To solve this question, you must use the formula: q=mc(change in temperature), where q is heat, m is mass, C is specific heat and temperature change is temperature change. The specific heat for ice is 2.1kJ/Kg x K (given). The change in temperature is 15 degrees Celsius (which you should change to kelvins so you can cancel out units), or 273 + 15 = 288K. The mass is 150 grams, which is 0.15 kg. Now, we can solve for q, heat. We will do this by substituting variables into the formula. After simplifying and cancelling out units, the answer we get is: 90.72kJ.
4 0
3 years ago
Select The on that most applys<br><br><br> Will mark brainliest
zepelin [54]

Answer:

A and B

Explanation:

3 0
2 years ago
0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium
Free_Kalibri [48]

Moles of potassium permanganate = 0.0008

<h3>Further explanation  </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range  

Reaction

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) +  10CO2(g) + 8H2O(1)

The end point ⇒titrant and analyte moles equal

titrant : potassium  permanganate-KMnO4

analyte : sodium oxalate - Na2C2O4

so moles of KMnO4 = moles of Na2C2O4

moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.264}{134 g/mol}\\\\mol=0.002

From equation, mol ratio  Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

\tt \dfrac{2}{5}\times 0.002=0.0008

6 0
3 years ago
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