The rate law for this reaction is [A]².
Balanced chemical reaction used in this experiment: A + B → P
The reaction rate is the speed at which reactants are converted into products.
Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.
Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².
More info about rate law: brainly.com/question/16981791
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These problems are a bit interesting. :)
First let's write the molecular formula for ammonium carbonate.
NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)
17.6 gNH4CO3
Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.
17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)
Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)
NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol
17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4
Now just take the molar mass we found to convert that amount into moles!
4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4
False, because water vapor, water vapour or aqueous vapor, is the gaseous phase of water. It is one state of water within the hydrosphere. Water vapor can be produced from the evaporation or boiling of liquid water or from the sublimation of ice. Unlike other forms of water, water vapor is invisible.
It would be considered acid rain in the sense of dew fall
Answer:
ΔG° = -533.64 kJ
Explanation:
Let's consider the following reaction.
Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)
The standard Gibbs free energy (ΔG°) can be calculated using the following expression:
ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)
where,
ni are the moles of reactants and products
ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products
ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)
ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)
ΔG° = -533.64 kJ
