Ionic bonds consist of a nonmetal and a metal, however, NH4 is a polyatomic, which makes it an exception. It can form ionic salt which is ionic in nature, so combined with a nonmetal, it becomes ionic.
One part of evidence that the color of the flame created is
from the metal ion and not from the chemical is that not any of the flames with
dissimilar metals had the similar color (for each metal had its own flame
color). Even if most of the metals tested had chloride, the colors of the flames
were all dissimilar. The two flames that both had copper (one had copper (II)
chloride and the other had copper (II) sulfate) were exactly close in color. The
one was green-blue, and the other was a bright green. This displays that they
were nearly the same, and the minor difference could be credited to error.
As if the energy isn't burned off it turns into fat.
A word equation is one that is written in words rather than using the normal symbols of the elements that are involved.
<h3>What is a word equation?</h3>
A word equation is one that is written in words rather than using the normal symbols of the elements that are involved. In this case we are asked to write the word equation for certain reactions and the states of matter of each of the terms in the reaction must be stated.
The word equations are as follows;
- sodium carbonate solution(aq) + silver nitrate solution(aq) ----> sodium nitrate(aq) + silver carbonate(aq)
- potassium carbonate solution(aq) + nitric acid (hydrogen nitrate) solution(aq) -----> potassium nitrate(aq) + water(l) + carbon dioxide
- barium nitrate solution (aq)+ potassium chromate solution(aq) -----> potassium nitrate (aq) + barium chromate (s)
Learn more about reaction equation:brainly.com/question/3588292
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Answer:
ΔH°rxn = -827.5 kJ
Explanation:
Let's consider the following balanced equation.
2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)
We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]
ΔH°rxn = -827.5 kJ
