The final temperature is -138 °C.
Explanation:
Using the equation of specific heat
We can easily find the final temperature of a 73.174 g of copper sample. As we know that specific heat is the amount of energy required to raise the temperature of the object to 1°C.
The specific heat of copper is known as 0.387 J/g°C and the initial temperature is said as 102 °C . The mass is given as 73.174 g. The heat released is 6800 J.
Since the heat is released the Q value will be negative.
Thus, the final temperature is -138 °C.
Answer:
E = 1.2443*10⁶ N/C
Explanation:
R = 41.6 cm = 0.416 m
Q₁ = 8.55 μC = 8.55*10⁻⁶C
Q₀ = 4.43 μC = 4.43*10⁻⁶C
r = 17.9 cm = 0.179 m
K = 9*10⁹ N*m²/C²
Since r < R we can apply Gauss's Law as follows
E = K*Q₀ / r²
⇒ E = (9*10⁹ N*m²/C²)*(4.43*10⁻⁶C) / (0.179 m)²
⇒ E = 1.2443*10⁶ N/C
An ion-dipole force is a type of intermolecular force in which forces of attraction or repulsion occur between neighboring ions, molecules or atoms.
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.
