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vodka [1.7K]
3 years ago
15

Which measure of a gas does the expression nRT represent? (1 point)

Chemistry
1 answer:
DaniilM [7]3 years ago
7 0
The Ideal Gas Equation
The term pVnRT is also called the compression factor and is a measure of the ideality of the gas. An ideal gas will always equal 1 when plugged into this equation……
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Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%, respectively)
Reika [66]

Answer:

Average atomic mass = 79.9034 amu

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

<u>For first isotope: </u>

% = 50.69 %

Mass = 78.9183 amu

<u>For second isotope: </u>

% = 49.31 %

Mass = 80.9163 amu

Thus,  

Average\ atomic\ mass=\frac{50.69}{100}\times {78.9183}+\frac{49.31}{100}\times {80.9163}\ amu

Average\ atomic\ mass=40.0036+39.8998\ amu

<u>Average atomic mass = 79.9034 amu</u>

4 0
3 years ago
Harry has a sample of an unknown material. The sample has a mass of 81 grams and a volume of 0.9 cubic centimeters. What is the
Salsk061 [2.6K]
Density = mass/volume
density = 81 grams/0.9 cubic centimetersdensity = 90 grams per cubic centimeter
The density of the sample is 90 grams per cubic centimeter.
8 0
3 years ago
The only way to break the bonds in a compound is with a
Sveta_85 [38]

Answer:

I think chemical reaction

Explanation:

8 0
3 years ago
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What is the molar mass of copper (Cu)
faust18 [17]

Answer:

63.546 u

Would be the molar mass of copper!

4 0
3 years ago
Please consider 1 kg of NH3 that was compressed from 2.5 bar and 30°C to 5.0 bar in a well-insulated compressor. Please determin
natali 33 [55]

Explanation:

(a)   The given data is as follows.

               mass = 1 kg = 1000 g        (as 1 kg = 1000 g)

          Molar mass of NH_{3} = 17 g/mol

           P_{1} = 2.5 bar = 2.5 \times 10^{5} Pa    (as 1 bar = 10^{5} Pa)

            P_{2} = 5 bar = 5 \times 10^{5} Pa

          T_{1} = 30^{o}C = 30 + 273 = 303 K

For adiabatic process,   PV^{\gamma} = constant = k

             \gamma = 1.33 = \frac{C_{p}}{C_{v}}        

              P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}

                (\frac{V_{2}}{V_{1}})^{\gamma} = \frac{P_{1}}{P_{2}}

        \frac{V_{2}}{V_{1}} = (\frac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}

    V_{2} = (\frac{2.5 \times 10^{5}}{5 \times 10^{5}})^{\frac{1}{1.33}} \times \frac{nRT_{1}}{P_{1}}            (as PV = nRT)

                 = (\frac{2.5}{5})^{\frac{1}{1.33}} \times \frac{58.82 \times 8.314 J/mol K \times 303 K}{2.5 \times 10^{5}}

                = 0.352 m^{3}

Also,       w = \frac{P_{1}V_{1} - P_{2}V_{2}}{\gamma - 1}

                  = \frac{2.5 \times 10^{5} \times 0.5927 - 5 \times 10^{5} \times 0.352}{1.33 - 1}

                  = -84318.2 J

As 1 kJ = 1000 J. So, -84318.2 J = -84.318 kJ

Hence, the work required in kJ is -84318.2 J.

(b)    It is known that for adiabatic system Q = 0,

                           \Delta U = Q - w

                          nC_{v}dT = -w

                             dT = \frac{-w}{nC_{v}}

                                  = \frac{84318.2}{58.82 \times 1.6}

                                  = 895.93 K

We known that dT = T_{1} - T_{2}

so,                   895.93 = 303 K - T_{2}                            

                         T_{2} = (895.93 - 303)K

                                     = 592.93 K

                                     = (592.93 - 273.15)^{o}C

                                     = 319.78^{o}C

Hence, the final temperature is 319.78^{o}C.

8 0
3 years ago
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