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3 years ago

Write the rate law for the following elementary reaction:

Chemistry

1 answer:

Harrizon [31]3 years ago

6 0

Answer:

Rate law = k1 [N2O5]^2

Explanation:

Rate law only cares about reactants and rate law can only be determined experimentally or by using the coefficients of reactants in elementary reactions

If 2A---->B is an elementary reaction... rate law for this is rate=k[A]^2

So if we look at your question...2N2O5---->2N2O4 + 02 (I think you are missing O2 in question because otherwise equation is unbalanced)

Rate law = k1 [N2O5]^2

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The second on it true

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3 years ago

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Does the identity of gas matter when predicting its behavior why or why not?

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3 years ago

Calculate the pOH of a solution if the concentration of hydroxide ions (OH-) is 1.9 x 10-5M?

OLEGan [10]

Answer:

9.28

Explanation:

pOH refers to a measure of hydroxide ions concentration. pOH tells about the alkalinity of a solution. If pOH is less than 7 then aqueous solutions are alkaline, acidic if pOH is greater than 7 and neutral if pOH is equal to 7.

Concentration of the hydroxide ions = 1.9 x 10-5 M

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pOH = 14 - pH

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6 0

3 years ago

The temperature at which a liquid becomes a solid is the:

Olenka [21]

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3 0

3 years ago

A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c

gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with this equation, and assuming no heat is lost to the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= 135,583 J = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

Now that we have the heat of combustion, we need to calculate the molar heat.

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write the conversion factor as 135.58 kJ/ 4.40 g.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

3 0

3 years ago