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3 years ago

Describe the molecular structure of water and explain why the water molecule is polar

Chemistry

1 answer:

pentagon [3]3 years ago

8 0

Explanation:

Hybridization of O in $H_2O = sp^3$

So, water molecule has four hybrid orbitals.

Two hybrid orbitals form 2 sigma bond with two H atoms.

Remaining two hybrid orbitals are occupied by two lone pairs.

Because of lone pair-lone pair repulsion, shape of $H_2O$ becomes bent.

Water molecule is polar because of difference in eletronegativities of O and H.

O is more electronegative as comapared to hydrogen. So bonding electrons get attracted towards O atom which results in the development of partial negative charge on O atom and partial positive charge on H atoms.

Because of development of partial negative and partial positive charge, water molecule becomes polar.

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3 years ago

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The transition elements or transition metals occupy the short columns in the center of the periodic table, between Group 2A and Group 3A.Explanation:

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4 years ago

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

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

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

3 0

3 years ago

The solubility of water in diethyl ether has been reported to be 1.468 % by mass.' Assuming that 23.0 mL of diethyl ether were a

melomori [17]

Explanation:

As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.

Hence, amount of diethyl ether present will be calculated as follows.

(100ml - 1.468 ml)

= 98.532 ml

So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.

Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.

Amount of water = $\frac{1.468 ml \times 23 ml}{98.532 ml}$

= 0.3427 ml

Now, when magnesium dissolves in water then the reaction will be as follows.

$Mg + H_{2}O \rightarrow Mg(OH)_{2}$

Molar mass of Mg = 24.305 g

Molar mass of $H_{2}O$ = 18 g

Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.

Amount of Mg = $\frac{24.305 g \times 0.3427 ml}{18 g}$

= 0.462 g

6 0

3 years ago