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Alekssandra [29.7K]
3 years ago
6

It’s easier to determine the elecron configurations for the p-block elements in periods 1,2,3 than to determine the electrons co

nfigurations for the rest of the p-block elements in the periodic table beacause
Chemistry
2 answers:
Scorpion4ik [409]3 years ago
5 0

Answer:It’s easier to determine the elecron configurations for the p-block elements in periods 1,2,3 than to determine the electrons configurations for the rest of the p-block elements in the periodic table beacause from period 4, specifically from the element 31 (Ga), the atoms start to fill the d orbitals, and the energy levels of the 3d orbitals ara quite similar to the energy levels of 4p orbitals. So, for the elements Cr and Cu the right configurations do not match the configurations predicted using Aufbau method and Hund rules. Those are not the only exceptions but the two first. All is due to the proximity of the energy of the d and p orbitals and the fact that the rearrangement of the electrons result in a lower energy

Explanation:

kotegsom [21]3 years ago
3 0
<span>It’s easier to determine the elecron configurations for the p-block elements in periods 1,2,3 than to determine the electrons configurations for the rest of the p-block elements in the periodic table beacause from period 4, specifically from the element 31 (Ga), the atoms start to fill the d orbitals, and the energy levels of the 3d orbitals ara quite similar to the energy levels of 4p orbitals. So, for the elements Cr and Cu the right configurations do not match the configurations predicted using Aufbau method and Hund rules. Those are not the only exceptions but the two first. All is due to the proximity of the energy of the d and p orbitals and the fact that the rearrangement of the electrons result in a lower energy level. </span>
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Answer:

a. 0.60

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Explanation:

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What did the immigration act of 1965 do?
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Changed the way quotas were allocated by ending the National Origins Formula that had been in place in the United States since the Emergency Quota Act<span> of 1921.</span>
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3 years ago
In pea plants, yellow seeds are dominant to green seeds. A heterozygous yellow-
san4es73 [151]

Answer:

The phenotypic percentage of having yellow or green seeds is 50% for having either of the two colours

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Thus, <u>the phenotypic percentage of having yellow or green seeds is 50% for having either of the two colours</u>.

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Hydrogen gas and nitrogen gas come together to form ammonia. If 25.0 g of nitrogen is mixed with hydrogen, what volume of hydrog
jeka94

Answer:

60.0L of hydrogen are needed

Explanation:

Based on the reaction:

3H₂ + N₂ ⇄ 2NH₃

<em>3 moles of hydrogen react 1 mole of nitrogen.</em>

<em />

To solve this question we have to find the moles of nitrogen. With the moles of nitrogen we can find the moles of hydrogen. Using PV = nRT at STP conditions we can find the volume as follows:

<em>Moles Nitrogen -Molar mass: 28g/mol-</em>

25.0g N₂ * (1mol / 28g) = 0.893 moles N₂

<em>Moles hydrogen: </em>

0.893 moles N₂ * (3mol H₂ / 1mol N₂) = 2.679 moles H₂

<em>Volume hydrogen:</em>

PV = nRT

V = nRT / P

<em>Where V is volume in liters,</em>

<em>n are moles of the gas: 2.679 moles</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 273.15K at STP</em>

<em>P is 1atm at STP</em>

Replacing:

V = 2.679mol*0.082atmL/molK*273.15K / 1atm

V =

<h3>60.0L of hydrogen are needed</h3>

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6 0
3 years ago
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