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mash [69]
3 years ago
8

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

mechanics works just as it does in our universe, except that there are seven d 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 L.
Chemistry
1 answer:
NARA [144]3 years ago
7 0

Answer:

Iron; [Ar] 3d6 4s2

Cobalt; [Ar] 3d6 4s2 4p1

Explanation:

We know that in our universe, there are five d orbitals known. However, in this strange universe L, there are only three d orbitals known.

The sixth and seventh first row transition elements are iron and cobalt. In universe L, the electronic configuration of iron and cobalt will be written as;

Iron; [Ar] 3d6 4s2

Cobalt; [Ar] 3d6 4s2 4p1

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What two atoms are held together by a covalent bond in NaHCO3
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Answer: covalent bond is the force of attraction that holds together two atoms that share a pair of valence electrons. Covalent bonds form only between atoms of nonmetals. The two atoms that are held together in a covalent bond may be atoms of the same element or different elements.

Explanation:

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Which of the following is the best definition of a physical change?
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I think answer is

C. Something that can be observed or measured while changing the identity of the substance
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Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

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5 0
2 years ago
A chemist prepares a solution of silver(I) nitrate by measuring out of silver(I) nitrate into a volumetric flask and filling the
leva [86]

Answer:

3.33 M

Explanation:

It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:

" <em>A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.</em>

<em>Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits.</em> "

In this case, first we <u>calculate the moles of AgNO₃</u>, using its molecular weight:

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Then we<u> convert the 150 mL of the volumetric flask into L</u>:

  • 150 / 1000 = 0.150 L

Finally we <u>divide the moles by the volume</u>:

  • 0.500 mol AgNO₃ / 0.150 L = 3.33 M
4 0
3 years ago
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