Answer:
Identify each equation as a composition reaction, a decomposition reaction, or neither.
Fe2O3 + 3 SO3 → Fe2(SO4)3
NaCl + AgNO3 → AgCl + NaNO3
(NH4)2Cr2O7 → Cr2O3 + 4 H2O + N2
Solution
In this equation, two substances combine to make a single substance. This is a composition reaction.
Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.
A single substance reacts to make multiple substances. This is a decomposition reaction.
Test Yourself
Identify the equation as a composition reaction, a decomposition reaction, or neither.
C3H8 → C3H4 + 2 H2
Explanation:
I hope I help :)))
Gas, as the particles have the most energy, and thus move the most.
Answer:C The final product cannot be converted back to the original ingredients.
Answer: well 18°C is really cool the mass of the water sample would be 28242 yeah that's right
Explanation:
The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).
The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.
The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.
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