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Nata [24]
3 years ago
14

Each atom of protium has one proton, no neutrons, and one electron.Each atom of deutrium has one proton, two neutrons, and one e

lectron. Are these the same or different elements? Why?
Chemistry
1 answer:
Ket [755]3 years ago
3 0

Answer:

These are the isotopes of same element. i.e. hydrogen.

Explanation:

Protium and deutrium are the isotopes of hydrogen.

Isotope:

Atoms of same elements can have different atomic mass but same atomic number . These atom of an elements are called isotopes.

Hydrogen consist of three stable isotopes protium, deutrium  and tritium.

These three isotopes are represented as H¹₁ , H²₁ and H³₁ respectively.

Protium consist of one proton and one electron while deutrium consist of one proton , one neutron and one electron. The number of neutron and proton is called mass number while number of electron or proton is called atomic number. The number of proton and electron are always same.That's why protium and deutrium have same atomic number but different atomic mass because there is no neutron present in protium that's why its atomic mass is less than by one from deutrium.

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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.

Heat_{\text{absorbed}}=Heat_{\text{released}}

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:

46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

c_3=30.68J/g^oC

Hence, the specific heat of calorimeter is 30.68 J/g°C

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