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3 years ago

10

What causes water molecules to be attracted to each other?

2 answers:

Fiesta28 [93]3 years ago

8 0

This polarity makes water molecules attracted to each other. The oxygen-hydrogen bond in the alcohol molecule is also polar. But, the carbon hydrogen bonds in the rest of the alcohol molecule are nonpolar. In these bonds, the electrons are shared more or less evenly.

dimulka [17.4K]3 years ago

7 0

Answer:

Both polar covalent bonds inside each water molecule and hydrogen bonds inside each water molecule are liable for attraction between water molecules.

Explanation:

Water is a polar protic molecule.
Due to large difference in electronegativity between oxygen and hydrogen atom, two O-H bonds in water molecule are polar.
Due to the presence of polar covalent bonds in water molecule, water molecules attract each other through oppositely polarized ends.
In water molecules, H atom is attached with highly electronegative oxygen atom. Hence hydrogen bonding interaction arises between water molecules.
So, both option (A) and (B) are correct

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4 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

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Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

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

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

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

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

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