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ELEN [110]
3 years ago
8

What intermolecular forces can occur between a CO3 ion and H2O molecules? List them

Chemistry
1 answer:
Orlov [11]3 years ago
5 0

Answer:

The intermolecular forces between CO3^2- and H2O molecules are;

1) London dispersion forces

2) ion-dipole interaction

3) hydrogen bonding

Explanation:

Intermolecular forces are forces of attraction that exits between molecules. These forces are weaker in comparison to the intramolecular forces, such as the covalent or ionic bonds between atoms in a molecule.

Considering CO3^2- and H2O, we must remember that hydrogen bonds occur whenever hydrogen is bonded to a highly electronegative atom such as oxygen. The carbonate ion is a hydrogen bond acceptor.

Also, the London dispersion forces are present in all molecules and is the first intermolecular interaction in molecular substance. Lastly, ion-dipole interactions exists between water and the carbonate ion.

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Because both the bonding electrons come from the oxygen atom. Explanation: A coordinate covalent bond is formed when both the bonding electrons are coming from the same atom

Explanation:

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370 cm3 of water at 80°C is mixed with 120 cm3 of water at 4°C. Calculate the final equilibrium temperature, assuming no heat is
NikAS [45]

Answer : The final temperature of the mixture is 61.4^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)

where,

c_1 = c_2 = specific heat of water = same

m_1 = m_2 = mass of water  =  same

\rho_1 = \rho_2 = density of water = 1.0 g/mL

V_1 = volume of water at 80.0^oC  = 370cm^3=370mL

V_2 = volume of water at 4^oC  = 120cm^3=120mL

T_f = final temperature of mixture = ?

T_1 = initial temperature of water = 80.0^oC

T_2 = initial temperature of water = 4^oC

Now put all the given values in the above formula, we get:

(\rho_1\times V_1)\times (T_f-T_1)=-(\rho_2\times V_2)\times (T_f-T_2)

(1.0g/mL\times 370mL)\times (T_f-80.0)^oC=-(1.0g/mL\times 120mL)\times (T_f-4)^oC

T_f=61.4^oC

Therefore, the final temperature of the mixture is 61.4^oC

6 0
4 years ago
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