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alukav5142 [94]
3 years ago
9

Disappearance of NO in ...- Help!!!? The following data were collected for the rate of disappearance of NO in the reaction 2 NO(

g) + O2(g) ¨ 2 NO2(g).
Run: [NO](M) [O2](M) Initail Rate (M/s) 1 .0126 .0125 1.41 X 10^-2 2 .0252 .0125 5.64 X 10^-2 3 .0252 .0250 1.13 X 10^-1 (a) What is the rate law for the reaction? (b) What are the units of the rate constant? (c) What is the average value of the rate constant calculated from the three data sets?
(d) What is the rate of disappearance of NO when [NO] = 0.0750 M and [O2] = 0.0100 M?
(e) What is the rate of disappearance of O2 at the concentrations given in part (d)?
please explain!
Chemistry
1 answer:
DaniilM [7]3 years ago
4 0
Bsisbso sis s dis xid sis
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In part A, when you added water to the sodium acetate in the flask, some of the chemical dissolved into the water. Would you cal
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Answer: The solution is a SATURATED solution.

Explanation:

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From the question above, when water was added to the sodium acetate in the flask, SOME of the chemical dissolved into the water, meaning that some remained undissolved. This is because a given volume of water can only dissolve a certain amount of chemical in it at room temperature. If more chemical is added to such a solution, the chemical will remain undissolved. Such a chemical solution is said to be a SATURATED SOLUTION.

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4 0
3 years ago
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When 1.045 g of CaO is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 32.3 °C.
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Answer:

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Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

\Delta H=m\times C\times \Delta T

Where,  

\Delta H  is the enthalpy change

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass of CaO = 1.045 g

Specific heat = 4.18 J/g°C

\Delta T=32.3-25.0\ ^0C=7.3\ ^0C

So,  

\Delta H=1.045\times 4.18\times 7.3\ J=31.88713\ J

Also, 1 J = 0.001 kJ

So,  

\Delta H=0.03189\ kJ

Also, Molar mass of CaO = 56.0774 g/mol

Moles=\frac{Mass}{Molar\ mass}=\frac{1.045}{56.0774}\ mol=0.01863\ mol

Thus, Enthalpy change in kJ/mol is:-

\Delta H=\frac{0.03189}{0.01863}\ kJ/mol=1.71\ kJ/mol

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2 years ago
Which pair consists of a molecular formula and its corresponding empirical formula?
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