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Tanya [424]
2 years ago
13

The temperature of the carbon dioxide atmosphere near the surface of venus is 475°c. calculate the average kinetic energy per mo

le of carbon dioxide molecules on venus. 2520 j/mol 5920 j/mol 4150 j/mol 5920 kj/mol 9330 j/mol
Chemistry
1 answer:
harkovskaia [24]2 years ago
8 0
Answer: <span>9330 j/mol
</span>
The temperature of the gas is 475 ° Celcius which is equal to: 475 +273= 748 °K. The formula for kinetic energy of individual atoms would be

K= 3/2 * kB * T
If kB is 1.38 * 10^-23 J/K and 1 mol is made from 6.02*10^23 molecule, then the kinetic energy of 1 mol CO2 would be:
K= 3/2 * kB * T
K= 3/2 * 1.38 * 10^-23  * 748 * 6.02 *10^23 =9324 J/mol
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Answer:

3.3 mol

Explanation:

5.2 mol * 14.9 L / 23.5 L

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Which substance can be described as cations bonded together by mobile electrons?a. S8(s)b. Ag(s)c. SiO2(s)d. KCl(s)e. Kr(l)
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Answer:

Ag(s)

Explanation:

Metal atoms are held together by the metallic bond. The metallic bond consists of cations held together by a sea of mobile electrons.

Thus, the crystal of Ag(s) is composed of silver ions held together by a sea of mobile electrons. Electrostatic forces of attraction holds the crystal together.

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2 years ago
Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No
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Answer:

  • Molality = 0.066 m
  • Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The <em>freezing-point depression</em> is expressed by:

  • ΔT=Kf * m

We put the data given by the problem and <u>solve for m</u>:

  • 2.63 °C = 40°C·kg/mol * m
  • m = 0.06575 m

For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:

  • 0.06575 m = Moles reserpine / kg camphor
  • 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We<u> calculate moles of reserpine:</u>

  • 0.06575 m = Moles reserpine / 0.025 kg camphor
  • Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:

  • 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>

8 0
2 years ago
a) (1 point) Build anthracene, optimize its geometry and examine its structure. Describe its shape. b) (1 point) Measure the C-C
oksano4ka [1.4K]

Answer:

a) The structure of anthracene is planar with all the pi electrons delocalized in the structure to maintain aromaticity.

b) The C-C bond length in anthracene is about 140 pm with all the bond lengths being similar to each other.

The standard C-C bond length is 154 pm while standard C=C bond is about 134 pm. Therefore the bond length in anthracene is smaller than standard C-C bond length and longer than standard C=C bond length. This can be explained from the fact that the C-C bonds in anthracene has be mixed characteristics of single and double bond because of the delocalization of pi electrons over the whole structure. As a result, they are neither fully single nor fully double bond in nature. Hence the observed bond lengths.

c) This molecule is not flat. The N-atom is sp3 hybridized here and the H-atom attached to N will remain out of plane.

Explanation:

8 0
2 years ago
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

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Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

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According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
2 years ago
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