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ale4655 [162]
4 years ago
8

The order in which atomic orbitals are filled is based on what factor?

Chemistry
2 answers:
Deffense [45]4 years ago
8 0
The highest energy level available is based on the factor
lesya [120]4 years ago
5 0
Aufbau<span> means "building-up." Aufbau principles determine the order in which atomic orbitals are filled as the atomic number is increased. For the hydrogen atom, the order of increasing orbital energy is given by 1</span>s<span> < 2</span>s<span> = 2</span>p<span> < 3</span>s<span> = 3</span>p<span> = 3</span>d<span>, etc. The dependence of energy on </span>nalone leads to extensive degeneracy, which is however removed for orbitals in many-electron atoms. Thus 2s<span> lies below 2</span>p<span>, as already observed in helium.</span>
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How much heat is required to raise the temperature of 10.35g of CCl4 from 32.1°c to 56.4°c
Travka [436]
We are going to use this formula:

Q = M*C*ΔT

when Q is the heat required 

M is the mass of CCl4 = 10.35 g

C is the specific heat capacity of CCl4 = 0.874J/g.c

and ΔT the change in temperature = 56.4 - 32.1 °C =24.3 °C

∴ Q = 10.35g * 0.874 * 24.3 °C

       = 219.8 J
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What is the texture and appearance of oil
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Explanation:

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What is the approximate mass of 25 cm3 of silver, if the density is 10.5 g/cm3? A. 0.42 g B. 2.4 g C. 42 g D. 260 g
bulgar [2K]

A."0.42"

or


D. but this formula is weird we take the mass 25 multiply by 10.5 we get 262.5

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6 0
3 years ago
As a whole, plants tend to release more oxygen than carbon dioxide into the environment, but the ratios of gases differ at vario
marshall27 [118]

During the day, due to the existence of sunlight, the procedure of photosynthesis elevations over that of respiration. Photosynthesis uses carbon dioxide to produce carbohydrates removing oxygen as a byproduct. During the night, because there is no sunlight, the rate of photosynthesis decreases. This, therefore, drives the rate of respiration to surpass the speed of photosynthesis. In respiration, oxygen is consumed and carbon dioxide is dismissed.

<h3>What is photosynthesis?</h3>
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3 0
1 year ago
Nitrogen (N2) enters a well-insulated diffuser operating at steady state at 0.656 bar, 300 K with a velocity of 282 m/s. The inl
Tasya [4]

Answer:

  1. The exit temperature of the Nitrogen would be 331.4 K.
  2. The area at the exit of the diffuser would be 7*10^{-3} m^2.
  3. The rate of entropy production would be 0.

Explanation:

  1. First it is assumed that the diffuser works as a isentropic device. A isentropic device is such that the entropy at the inlet is equal that the entropy T the exit.
  2. It will be used the subscript <em>1 for the</em> <em>inlet conditions of the nitrogen</em>, and the subscript <em>2 for the exit conditions of the nitrogen</em>.
  3. It will be called: <em>v</em> the velocity of the nitrogen stream, <em>T</em> the nitrogen temperature, <em>V</em> the volumetric flow of the specific stream, <em>A</em> the area at the inlet or exit of the diffuser and, <em>P</em> the pressure of the nitrogen flow.
  4. It is known that <em>for a fluid flowing, its volumetric flow is obtain as:</em> V=v*A,
  5. Then for the inlet of the diffuser: V_1=v_1*A_1=282\frac{m}{s}*4.8*10^{-3}m^2=1.35\frac{m^3}{s}
  6. For an ideal gas working in an isentropic process, it follows that: \frac{T_{2} }{T_1}=(\frac{P_2}{P_1})^k where each variable is defined according with what was presented in step 2 and 3, and <em>k </em>is the heat values relationship, 1.4 for nitrogen.
  7. Then <em>solving</em> for T_2, the temperature of the nitrogen at the exit conditions: T_2=T_1(\frac{P_2}{P_1})^k then, T_2=300 K (\frac{0.9 bar}{0.656 bar})^{(\frac{1.4-1}{1.4})}=331.4 K
  8. Also, for an ideal gas working in an isentropic process, it follows that:  \frac{P_2}{P_1}= (\frac{V_1}{V_2})^k, where each variable is defined according with what was presented in step 2 and 3, and <em>k</em> is the heat values relationship, 1.4 for nitrogen.
  9. Then <em>solving</em> for V_2 the volumetric flow at the exit of the diffuser: V_2=V_1*\frac{1}{\sqrt[k]{\frac{P_2}{P_1}}}=\frac{1.35\frac{m^3}{s}}{\sqrt[1.4]{\frac{0.9bar}{0.656bar} }}=1.080\frac{m^3}{s}.
  10. Knowing that V_2=1.080\frac{m^3}{s}, it is possible to calculate the area at the exit of the diffuser, using the relationship presented in step 4, and solving for the required parameter: A_2=\frac{V_2}{v_2}=\frac{1.08\frac{m^3}{s} }{140\frac{m}{s}}=7.71*10^{-3}m^2.
  11. <em>To determine the rate of entropy production in the diffuser,</em> it is required to do a second law balance (entropy balance) in the control volume of the device. This balance is: S_1+S_{gen}-S_2=\Delta S_{system}, where: S_1 and S_2 are the entropy of the stream entering and leaving the control volume respectively, S_{gen} is the rate of entropy production and, \Delta S_{system} is the change of entropy of the system.
  12. If the diffuser is operating at stable state is assumed then \Delta S_{system}=0. Applying the entropy balance and solving the rate of entropy generation: S_{gen}=S_2-S_1.
  13. Finally, it was assume that the process is isentropic, it is: S_1=S_2, then S_{gen}=0.
6 0
3 years ago
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