Lead have:
82 electrons
82 protons
125 neutrons
Explanation:
The number of electrons and protons are given by the atomic number of the element.
Lead have 82 electrons and 82 protons.
The number of neutrons are given by the difference between the mass number and atomic number.
For lead number of neutrons = 207 - 82 = 125.
Learn more about:
particles in the atom
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Answer 1 : The balanced complete ionic equation will be,
![NaOH(aq)+HNO_3(aq)\rightarrow H_2O(l)+NaNO_3(aq)](https://tex.z-dn.net/?f=NaOH%28aq%29%2BHNO_3%28aq%29%5Crightarrow%20H_2O%28l%29%2BNaNO_3%28aq%29)
![Na^+(aq)+OH^-(aq)+H^+(aq)+NO_3^-(aq)\rightarrow H_2O(l)+Na^+(aq)+NO_3^-(aq)](https://tex.z-dn.net/?f=Na%5E%2B%28aq%29%2BOH%5E-%28aq%29%2BH%5E%2B%28aq%29%2BNO_3%5E-%28aq%29%5Crightarrow%20H_2O%28l%29%2BNa%5E%2B%28aq%29%2BNO_3%5E-%28aq%29)
By removing the spectator ion in this equation, we get the balanced ionic equation.
![OH^-(aq)+H^+(aq)\rightarrow H_2O(l)](https://tex.z-dn.net/?f=OH%5E-%28aq%29%2BH%5E%2B%28aq%29%5Crightarrow%20H_2O%28l%29)
Answer 2 : The balanced net ionic equation will be,
![2Na_3PO_4(aq)+3NiCl_2(aq)\rightarrow Ni_3(PO_4)_2(s)+6NaCl(aq)](https://tex.z-dn.net/?f=2Na_3PO_4%28aq%29%2B3NiCl_2%28aq%29%5Crightarrow%20Ni_3%28PO_4%29_2%28s%29%2B6NaCl%28aq%29)
![6Na^+(aq)+2PO_4^{3-}(aq)+3Ni^+(aq)+6Cl^-(aq)\rightarrow Ni_3(PO_4)_2(s)+6Na^+(aq)+6Cl^-(aq)](https://tex.z-dn.net/?f=6Na%5E%2B%28aq%29%2B2PO_4%5E%7B3-%7D%28aq%29%2B3Ni%5E%2B%28aq%29%2B6Cl%5E-%28aq%29%5Crightarrow%20Ni_3%28PO_4%29_2%28s%29%2B6Na%5E%2B%28aq%29%2B6Cl%5E-%28aq%29)
By removing the spectator ion in this equation, we get the balanced ionic equation.
![2PO_4^{3-}(aq)+3Ni^+(aq)\rightarrow Ni_3(PO_4)_2(s)](https://tex.z-dn.net/?f=2PO_4%5E%7B3-%7D%28aq%29%2B3Ni%5E%2B%28aq%29%5Crightarrow%20Ni_3%28PO_4%29_2%28s%29)
Answer 3 : The balanced net ionic equation will be,
![2Na_3PO_4(aq)+3NiCl_2(aq)\rightarrow Ni_3(PO_4)_2(s)+6NaCl(aq)](https://tex.z-dn.net/?f=2Na_3PO_4%28aq%29%2B3NiCl_2%28aq%29%5Crightarrow%20Ni_3%28PO_4%29_2%28s%29%2B6NaCl%28aq%29)
![6Na^+(aq)+2PO_4^{3-}(aq)+3Ni^+(aq)+6Cl^-(aq)\rightarrow Ni_3(PO_4)_2(s)+6Na^+(aq)+6Cl^-(aq)](https://tex.z-dn.net/?f=6Na%5E%2B%28aq%29%2B2PO_4%5E%7B3-%7D%28aq%29%2B3Ni%5E%2B%28aq%29%2B6Cl%5E-%28aq%29%5Crightarrow%20Ni_3%28PO_4%29_2%28s%29%2B6Na%5E%2B%28aq%29%2B6Cl%5E-%28aq%29)
By removing the spectator ion in this equation, we get the balanced ionic equation.
![2PO_4^{3-}(aq)+3Ni^+(aq)\rightarrow Ni_3(PO_4)_2(s)](https://tex.z-dn.net/?f=2PO_4%5E%7B3-%7D%28aq%29%2B3Ni%5E%2B%28aq%29%5Crightarrow%20Ni_3%28PO_4%29_2%28s%29)
Balanced equations : Balanced equations are the equations in which the number of individual elements present on the reactant side must be equal to the number of individual elements present on the product side.
Spectator ions : It is defined as the ions which do not participate in the chemical reaction. These ions exists in the same form on both the sides of the reaction.
Answer: the enthalpy of reaction is, -155 kJ
Explanation:-
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The final reaction is:
![\Delta H_{rxn}=?](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%3F)
The intermediate balanced chemical reaction will be,
(1)
![\Delta H_1=-284kJ](https://tex.z-dn.net/?f=%5CDelta%20H_1%3D-284kJ)
(2)
![\Delta H_2=-527kJ](https://tex.z-dn.net/?f=%5CDelta%20H_2%3D-527kJ)
(3)
![\Delta H_3=44.0kJ](https://tex.z-dn.net/?f=%5CDelta%20H_3%3D44.0kJ)
Now multiplying (3) by 2 and adding all the equations, we get :
![\Delta H_{rxn}=\Delta H_1+\Delta H_2+2\times \Delta H_3](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5CDelta%20H_1%2B%5CDelta%20H_2%2B2%5Ctimes%20%5CDelta%20H_3)
![\Delta H_{rxn}=(-284)+(-527)+2\times (44)](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%28-284%29%2B%28-527%29%2B2%5Ctimes%20%2844%29)
![\Delta H_{rxn}=-155kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D-155kJ)
Therefore, the enthalpy of reaction is, -155 kJ