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Bumek [7]
4 years ago
14

Part a choose the molecule or compound that exhibits dipole-dipole forces as its strongest intermolecular force. Choose the mole

cule or compound that exhibits dipole-dipole forces as its strongest intermolecular force. N2 h2 cbr4 so2 bcl3

Chemistry
1 answer:
mart [117]4 years ago
7 0

Answer :  SO_{2} is the molecule that exhibit dipole-dipole forces as its strongest intermolecular force.

Explanation : The shape or geometry of given molecules of N_{2}, H_{2}, CBr_{4}, BCl_{3} and SO_{2} are linear, linear, tetrahedral, trigonal planar and angular respectively.

The given molecules N_{2}, H_{2}, CBr_{4}, BCl_{3} are symmetrical molecules. These symmetric molecules cannot exists as dipole even they contain polar bonds. Thus, these molecules will not exhibit dipole-dipole forces.

SO_{2} molecule has polar bonds and unsymmetrical bonds as shown in attached image. This molecule cannot exhibit H-bonding. It exhibit only dipole-dipole forces.

Therefore, SO_{2} molecule exhibit dipole-dipole forces as strongest intermolecular force.

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Ferrous sulfate (FeSO4) was used in the
BARSIC [14]

Ferrous sulfate (FeSO4) is obtained by the reaction of aqueous ferric sulfate with gaseous sulfur dioxide and water according to the equation below:

Aqueous ferric sulfate = Fe_2(SO_4)_3_{(aq)}

Gaseous sulfur dioxide = SO_2_{(g)}

Water = H_2O_{(l)}

Aqueous sulfuric acid =  H_2SO_4_{ (aq)}

Aqueous ferrous sulfate =  FeSO_4_{ (aq)}

Thus, the balance equation becomes:

Fe_2(SO_4)_3_{(aq)} +  SO_2_{(g)} + 2H_2O_{(l)} ---> 2 H_2SO_4_{ (aq)} + 2 FeSO_4_{ (aq)}

More on ferrous sulfate can be found here:  brainly.com/question/10834028

6 0
3 years ago
A 4.0 g sample of iron was heated from 0°C to 20.°C. It absorbed 35.2 J of energy as heat. What is the specific heat of this pie
Crazy boy [7]

Answer:

Explanation:i joule is equal to 0.238902957619 calories so 1251 joules is equal to 298.87 calories divided by 25.0 degrees centigrade is equal to 11.95 calories divided by the 35.2 gram sample weight to get the calories per gram per degree centigrade would come to 0.3396 calories/gram degree centigrade. Presumably this, if correct, could be used to obtain the metal in question by consulting a chart or table with specific heats of various metals because they should always be the same specific heat for each metal.

5 0
3 years ago
Why do some elements that are classified as metals not look metallic?
Shkiper50 [21]

Answer:

So, that students like us get confuse!!!

Explanation:

6 0
3 years ago
Vitamin B information please​
Montano1993 [528]

B vitamins play a vital role in maintaining good health and well-being. As the building blocks of a healthy body, B vitamins have a direct impact on your energy levels, brain function, and cell metabolism.

Hope this helps; have a great day!

8 0
3 years ago
In an experiment, a student gently heated a hydrated copper compound to remove the water of hydration. The following data was re
Ronch [10]

The percent of water in the hydrated salt will be 55.08 %

<h3>What is mass?</h3>

Mass is a numerical measure of quantity, which is a basic feature of all matter. The kilogram is the international system of units of mass.

Given data;

Mass of crucible, cover, and contents before heating = 23.4 g.

Mass of empty crucible and cover = 18.82 g.

Mass of crucible, cover, and contents after heating to constant mass = 20.94 g.

⇒Mass of hydrated salt used =  Mass of the crucible, cover, and contents before heating - Mass of the empty crucible and cover                    

⇒Mass of hydrated salt used =   = 23.54 g – 18.82 g

⇒Mass of hydrated salt used =  = 4.72 g

⇒Mass of dehydrated salt after heating = Mass of the crucible, cover, and contents after heating to constant mass-Mass of the empty crucible and cover

⇒Mass of dehydrated salt after heating = 20.94 g – 18.82 g

⇒Mass of dehydrated salt after heating = 2.12 g

⇒ Mass of water liberated from salt = Mass of hydrated salt used -   Mass of dehydrated salt after heating

⇒  Mass of water liberated from salt = 4.72 g – 2.12 g    

⇒  Mass of water liberated from salt = 2.60 g  

   

Water % in the hydrated salt is found as;

% water =(mass of water/  mass of hydrated salt)× 100

% water =(2.60/4.72) × 100

% water =55.08 %

Hence the percent of water in the hydrated salt will be 55.08 %

To learn more about the mass, refer to the link.\

brainly.com/question/13073862

#SPJ1

8 0
2 years ago
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