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o-na [289]
3 years ago
10

Theobromine (greek theobroma, meaning "food of the gods") is a common constituent in coca. fill in all the lone pairs on each he

teroatom (atoms that are not carbon or hydrogen). what is the hybridization of each of these atoms? (1 point)

Chemistry
1 answer:
Ganezh [65]3 years ago
8 0

The lone pair of electrons in Theobromine are shown in RED color in attached figure.

Details:

           In chemistry there are two types of electrons.

i) Bonding Pair Electrons:

                                        These are those electrons which are being shared and are involved in making bonds. They are also called as sharing electrons. In given structure all the solid bonds either single or double are made up of bonding pair electrons.

ii) Lone Pair / Non-Bonding Pair Electrons:

                                                                     Those electrons which doesn't take part in bonding and are not shared between atoms. In given structure the lone pair of electrons are found only on nitrogen atoms (single lone pair) and oxygen atoms (two lone pair of electrons) respectively.

Hybridization of Nitrogen Atoms;

                                                      N₁  =  Sp²

                                                      N₂  =  Sp³

                                                      N₃  =  Sp³

                                                      N₄  =  Sp³

Hybridization of Oxygen Atoms;

                                                      O₁  =  Sp²

                                                      O₂  =  Sp²

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<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

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The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 46.7 g

m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

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c_3=30.68J/g^oC

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