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2 years ago

1. An atom of an elephant has atomic mass of 39 and has number of neutrons as 20 calculate protons number.

Chemistry

1 answer:

r-ruslan [8.4K]2 years ago

5 0

Answer:

See Explanation

Explanation:

1) Let us recall that;

Mass number = number of protons + number of neutrons

Where;

Number of protons = ?

Number of neutrons =20

Mass number =39

Number of protons = mass number - number of neutrons

Mass of protons =39 - 20= 19 protons

2) Atomic number = 13

Number of neutrons =14

Then the mass number =13 +14 =27

The symbol of the element is 27X13

The number of protons= the number of electrons =13

The atomic number is the same as the proton number =13

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Answer:

The enthalpy of the reaction is coming out to be -380.16 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]$

We are given:

$\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ$

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

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Explanation:

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Answer:

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