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dlinn [17]
3 years ago
11

One kilogram of saturated steam at 373 K and 1.01325 bar is contained in a rigid walled vessel. It has a volume of 1.673 m3. It

is cooled to a temperature at which the specific volume of water vapour is 1.789 m. The amount of water vapour condensed in kilograms is (a) 0.0 (b) 0.065 (c) 0.1 (d) 1.0
Chemistry
1 answer:
NeTakaya3 years ago
8 0

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

        Initial volume (v_{1}) = 1.673 m^{3}

          Final volume (v_{2}) = 1.789 m^{3}

As, the amount of water vapor condensed will be as follows.

                     \frac{(v_{2} - v_{1})}{v_{2}}

                     = \frac{(1.789 m^{3} - 1.673 m^{3})}{1.789 m^{3}}

                     = \frac{0.116 m^{3}}{1.789 m^{3}}

                     = 0.065 kg

Hence, we can conclude that the amount of water vapour condensed in kilograms is 0.065 kg.

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A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of oxygen gas, producing 9.58
Scilla [17]

Answer:

\% O=27.6\%

Explanation:

Hello,

In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:

- Moles of carbon are contained in the 9.582 grams of carbon dioxide:

n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}  =0.218molC

- Moles of hydrogen are contained in the 3.922 grams of water:

n_H=3.922gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.436molH

- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:

m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO

Finally, we compute the percent by mass of oxygen:

\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%

Regards.

5 0
3 years ago
How do alpha particles compare to gamma rays?
Fofino [41]

Answer:

Unlike alpha and beta particles, which have both energy and mass, gamma rays are pure energy. Gamma rays are similar to visible light, but have much higher energy. Gamma rays are often emitted along with alpha or beta particles during radioactive decay.

-Radiation basics.

Explanation:

5 0
3 years ago
Read 2 more answers
Use the specific heat capacity that you calculated for granite to determine how many grams of granite at the initial temperature
Valentin [98]

Answer:

Explanation:

specific heat of granite s = .79 J / g / k

let the mass of granite = m

heat lost by granite = heat gained by water

heat lost = mass x specific heat x drop in temperature

= m x .79 x (80 - 20.45)

heat gained by water

= 3000 x 4.186 x (20.45- 20)

heat lost by granite = heat gained by water

m x .79 x 59.55  =  3000 x 4.186 x .45

m = 120.12 g .

4 0
3 years ago
Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr
Alecsey [184]

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}, where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

5 0
3 years ago
What is the entropy change of the surroundings
KiRa [710]

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

Amount of Acetone given = 10.8 g

Number of moles is calculated by using the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of acetone = 58 g/mol

Number of moles = \frac{10.8}{58}=0.1862moles

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = \frac{32.3}{1}\times 0.1862=5.828kJ/mol

To calculate the entropy change for the system, we use the formula:

\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol   (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

\Delta S_{system}+\Delta S_{surrounding}=0

\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol

5 0
3 years ago
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