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Umnica [9.8K]
3 years ago
8

Values for the molar mass of nitrogen, oxygen, and nitrogen dioxide molecules are given in the table below. What mass of nitroge

n dioxide molecules is formed when 1 mole of nitrogen reacts with 2 moles of oxygen and forms nitrogen dioxide? Choices: A). 92.02g B). 46.01g C). 23.00g D) 2.00g
Chemistry
1 answer:
Gnoma [55]3 years ago
4 0

Answer:

A). 92.02g

Explanation:

Equation of the reaction;

N2 (g)+ 2O2(g)------> 2NO2(g)

Note that the balanced reaction equation is the first step in solving any problem on stoichiometry. Once the reaction equation is correct, the question can be easily solved.

Reaction of one mole of nitrogen gas with two moles of oxygen gas yields two moles of nitrogen dioxide.

Mass of two moles of nitrogen dioxide= 2[14 + 2(16)] = 2[14+32]= 2[46]= 92 gmol-1

Therefore; Mass of two moles of nitrogen dioxide is 92

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3 years ago
An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr
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<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

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Rate of flow of ideal gas , n = 4 kmol/hr = \frac{4\times 1000mol}{3600s}=1.11mol/s    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

We know that:

\Delta U=0   (For isothermal process)

So, by applying first law of thermodynamics:

\Delta U=\Delta q-\Delta W

\Delta q=\Delta W      .......(1)

Now, calculating the work done for isothermal process, we use the equation:

\Delta W=nRT\ln (\frac{P_1}{P_2})

where,

\Delta W = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

P_1 = initial pressure = 100 kPa

P_2 = final pressure = 50 kPa

Putting values in above equation, we get:

\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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Some examples of compounds are H₂O, NaCl, H₂O₂, CH₃COOH. As you see, they have a chemical formula which states the kind and number of the atoms that form them.


They are different to mixtures, which are formed by two or more compounds, in a variable proportion, and can be separated by physical media. Some examples of mixtures are the solutions (e.g. NaCl dissolved in H₂O), and some solid mixtures (e.g. a mixture of marbles and sand).

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3 years ago
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