Answer: its a reaction
Explanation:
like coke its a reaction between gas and oxygen
The molar mass of methylammonium bromide is 111u.
<h3>What is molar mass?</h3>
The molar mass is defined as the mass per unit amount of substance of a given chemical entity.
Multiply the atomic weight (from the periodic table) of each element by the number of atoms of that element present in the compound.
Add it all together and put units of grams/mole after the number.
Atomic weight of H is 1u
Atomic weight of N is 14u
Atomic weight of C is 12u
Atomic weight of Br is 79u
Calculating molar mass of 
=2(1 x3+ 14+12+ 1 x 3 +79) = 111u
Hence, the molar mass of methylammonium bromide is 111u.
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2H₂(g) + O₂(g) ⇄ 2H₂O(l)
Δngas = 0 - (2 +1)
= -3
<h3>
What is Δngas?</h3>
The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.
Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.
<h3>
Definition of equilibrium</h3>
When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.
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Answer:
Explanation:
In a nuclear reaction, the total mass and total atomic number remains the same.
Am has an atomic number of 95. So correct reaction is:-
To calculate A:
Total mass on reactant side = total mass on product side
343 + 4 = A + 2
A = 345
To calculate Z:
Total atomic number on reactant side = total atomic number on product side
95 + 2 = Z + 0
Z = 97
Hence, the isotopic symbol of unknown element is
Answer:
Hi! In this case, the reactioncan be correctly balance according to this: 2Al(s) + 3CuSO4(aq) –> Al2(SO4)3(aq) + 3Cu(s).
Explanation:
In this particulary reaction,two semi-reactions happens.
One involving the metallic aluminum that suffers an oxidation reaction:
Al (s) -> Al3 + (aq) + 3e–
and another is a reduction reaction involving copper;
2e– + Cu2 + (aq) -> Cu (s)
