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tigry1 [53]
3 years ago
10

In Universe , recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quant

um mechanics works just as it does in our universe, except that there are seven orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the sixth and seventh elements in the first transition series in Universe . Note: you may use to stand for the electron configuration of the noble gas at the end of the row before the first transition series.
Chemistry
1 answer:
lawyer [7]3 years ago
8 0

Answer:

According to Hund's rule and the Aufbau principle in which the orbitals must be filled with electrons, they are not strictly applied in the real universe, because the intermediate and electron-filled atomic orbitals are very stable . Because there are four d-orbitals in universe L, a typical half-full configuration will be xd4 and its full configuration will be xd8, where x is the primary orbital for any specific element. Here is an example:

Vahadium ₂₃V

in real universe: [Ar]₈ 3d³4s²

in universe L: [Ar]₁₈ 3d⁴4s¹

Chromium

in real universe: [Ar]₈ 3d⁵4s¹

in universe L: [Ar]₁₈ 3d⁴4s²

Explanation:

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wolverine [178]
The first one i think
8 0
3 years ago
Combustion of hydrocarbons such as dodecane (C12H26) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth'
miss Akunina [59]

Answer:

A. 2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

B. 761.42 L

Explanation:

A. Step 1:

The equation for the reaction.

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

A. Step 2:

Balancing the equation.

The equation can be balance as follow:

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

There are 12 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 12 in front of CO2 as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + H2O(g)

There are 26 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 13 in front of H2O as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + 13H2O(g)

Now, there are a total of 37 atoms of O2 on the right side and 2 atoms on the left. It can be balance by putting 37/2 in front of O2 as illustrated below:

C12H26(l) + 37/2O2(g) —> 12CO2(g) + 13H2O(g)

Multiply through by 2 to clear the fraction from the equation.

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

Now the equation is balanced

B. Step 1:

We'll by obtaining the number of mole of C12H26 in 0.450 kg of C12H26. This is illustrated below:

Molar Mass of C12H26 = (12x12) + (26x1) = 144 + 26 = 170g/mol

Mass of C12H26 = 0.450 kg = 0.450x1000 = 450g

Number of mole of C12H26 =?

Number of mole = Mass/Molar Mass

Number of mole of C12H26 = 450/170

Number of mole of C12H26 = 2.65 moles

B. Step 2:

Determination of the number of mole of CO2 produced by the reaction. This is illustrated below:

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

From the balanced equation above,

2 moles of C12H26 produced 24 moles of CO2.

Therefore, 2.65 moles of C12H26 will produce = (2.65x24)/2 = 31.8 moles of CO2.

B. Step 3:

Determination of the volume of CO2 produced by the reaction.

Pressure (P) = 1 atm

Temperature (T) = 19°C = 19°C + 273 = 292K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) = 31.8 moles

Volume (V) =?

The volume of CO2 produced by the reaction can b obtained by applying the ideal gas equation as follow:

PV = nRT

1 x V = 31.8 x 0.082 x 292

V = 761.42 L

Therefore, the volume of CO2 produced is 761.42 L

5 0
3 years ago
Read 2 more answers
Calculate the mass in grams for each of the following liquids.
WITCHER [35]
The density is calculated as mass per volume, so if we want to solve for mass, we would multiply density by volume.
For Part A: if we have a density of 0.69 g/mL, and a volume of 280 mL, multiplying these will give a mass of: (0.69 g/mL)(280 mL) = 193.2 g. Rounded to 2 significant figures, this is 190 g gasoline.
For Part B: if we have a density of 0.79 g/mL, and a volume of 190 mL, multiplying these will give a mass of: (0.79 g/mL)(190 mL) = 150.1 g. Rounded to 2 significant figures, this is equal to 150 g ethanol.
3 0
3 years ago
The weight of an object ____.
miv72 [106K]

Explanation:

the. correct option is d

3 0
2 years ago
HELP PLEASE I HAVE A TEST TODAY AND I DON'T UNDERSTAND ANY OF THIS...
myrzilka [38]

Answer:

About 67 grams or 67.39 grams

Explanation:

First you would have to remember a few things:

 enthalpy to melt ice is called enthalpy of fusion.  this value is 6.02kJ/mol

  of ice  

 it takes 4.18 joules to raise 1 gram of liquid water 1 degree C

 water boils at 100 degrees C and water melts above 0 degrees C

 1 kilojoules is 1000 joules

  water's enthalpy of vaporization (steam) is 40.68 kJ/mol

  a mole of water is 18.02 grams

  we also have to assume the ice is at 0 degrees C

Step 1

Now start with your ice.  The enthalpy of fusion for ice is calculated with this formula:

q = n x ΔH    q= energy, n = moles of water, ΔH=enthalpy of fusion

Calculate how many moles of ice you have:

150g x (1 mol / 18.02 g) = 8.32 moles

Put that into the equation:

q = 8.32 mol x 6.02 = 50.09 kJ of energy to melt 150g of ice

Step 2

To raise 1 gram of water to the boiling point, it would take 4.18 joules times 100 (degrees C)  or 418 joules.

So if it takes 418 joules for just 1 gram of water, it would take 150 times that amount to raise 150g to 100 degrees C.  418 x 150 = 62,700 joules or 62.7 kilojoules.

So far you have already used 50.09 kJ to melt the ice and another 62.7 kJ to bring the water to boiling.  That's a total of 112.79 kJ.

Step 3

The final step is to see how much energy is left to vaporize the water.

Subtract the energy you used so far from what you were told you have.

265 kJ - 112.79 kJ = 152.21 kJ

Again q = mol x ΔH (vaporization)

You know you only have 152.21 kJ left so find out how many moles that will vaporize.

152.21 kJ = mol x 40.68  or   mol = 152.21 / 40.68  = 3.74 moles

This tells you that you have vaporized 3.74 moles with the energy you have left.

Convert that back to grams.

3.74 mol   x  ( 18.02 g / 1 mol ) = 67.39 grams

5 0
2 years ago
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