Answer:
N=2
H=6
Explanation:
1.Balance a chemical equation in terms of moles.
2.Use the balanced equation to construct conversion factors in terms of moles.
3.Calculate moles of one substance from moles of another substance using a balanced chemical equation.
The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products.
(P.s it could also be where you have to solve it in which you have to simplify it first then solve it.) like adding them all up.
Hope this is the answer. :)
When it’s warmer so when temperature encreases
Answer:
Number of protons = 82
Explanation:
Given data:
Atomic number of Pb = 82
Mass number of Pb-207 = 207
Number of protons = ?
Solution:
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.
Atomic number of Pb = number of protons or electrons
82 = number of protons or electrons
Mass number = Number of protons + number of neutron
207 = 82 + number of neutron
Number of neutrons = 207 - 82
Number of neutrons = 125
Here's the balanced equation for given Double displacement reaction ~
The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!