Which statement about the spectrophotometric analysis of Red 40 dye is TRUE?

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

Advocard [28]

Answer:

Manganese, Fifth transition element

[X] 3d⁶ 4s¹

Iron, Sixth transition element

[X] 3d⁶ 4s²

Explanation:

Complete Question

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quantum mechanics works as it does in our universe, except that there are six 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. Note; you may use [X] to stand for the electron configuration of the noble gas at the end of the row before the first transition series.

Solution

In our universe, there are 5 d orbitals.

And according to Aufbau's principles that electrons fill the lower energy orbitals before they fill higher energy orbitals and Hund's Rule that states that electrons are fed singly to all the orbitals of a subshell before pairing occurs.

The fifth and sixth transition elements in our universe is then Manganese and Iron respectively.

Manganese - [Ar] 3d⁵ 4s²

Iron - [Ar] 3d⁶ 4s²

So, in the new universe L, where there are six d orbitals, for manganese, the fifth transition metal, because half filled orbitals are more stable than partially filled orbitals (that woukd have been rhe case if we leave 5 electrons on the 3d orbital), the 4s orbital is filled to half of its capacity and the one electron removed from the 4s is used to fill the six 3d orbital to half of its capacity too.

For the sixth transition element, the new extra electron just fills the lower energy 4s orbital, leaving the six 3d orbitals all half-filled.

Hence, they both have ground state configurations of

- Manganese, Fifth transition element

[X] 3d⁶ 4s¹

- Iron, Sixth transition element

[X] 3d⁶ 4s²

Hope this Helps!!!

7 0

2 years ago

When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t

Harlamova29_29 [7]

Answer: Molarity of $MgCl_2$ in the original sample was 1.96M

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{\text{no of moles}}{\text{Volume in L}}$

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{186g}{95g/mol}=1.96$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.96}{1.00L}$

$Molarity=1.96mol/L$

Thus molarity of $MgCl_2$ in the original sample was 1.96M

4 0

2 years ago

Are the strengths of the interactions between the particles in the solute and between the particles in the solvent before the so

Colt1911 [192]

Answer:

Less than

Explanation:

The process of dissolution occurs as a kind of "tug of war". On one side are the solute-solute and solvent-solvent interaction forces, while on the other side are the solute-solvent forces.

Only when the solute-solvent forces are strong enough to overcome the pre-mixing forces do they overcome the "tug of war", and thus dissolution occurs.

Thus, it is concluded that the interaction forces between solute particles and solvent particles before they are combined are less than the interaction forces after dissolution.

5 0

2 years ago

A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav

dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of $_{19}^{40}\textrm{K}$ radioisotope follows:

$Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles$

$Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles$

$_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e$

By the stoichiometry of above reaction,

1 mole of $_{40}^{91}\textrm{Zr}$ is produced by 1 mole $_{41}^{91}\textrm{Nb}$

So, 0.17 moles of $_{40}^{91}\textrm{Zr}$ will be produced by = $\frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}$

Amount of $_{82}^{212}\textrm{K}$

decomposed will be = 0.17 moles

Initial amount of $_{40}^{91}\textrm{Nb}$ will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = 0.024

$a_o$ = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

$0.024=\frac{0.194}{2^n}$

$n=3$

Therefore, 3 half lives have passed.

3 0

2 years ago

Read 2 more answers

A hypothetical substance has a melting point of −10°C and a boiling point of 155°C. If this substance is heated from 2°C to the

igor_vitrenko [27]

Answer:

The specific heat capacity of liquid and the het of vaporization is used.

.

Explanation:

Step 1: Data given

A substance at temperature 2°C.

The substance has a melting point of −10°C and a boiling point of 155°C.

The initial temperature is 2°C which is between the melting point (-10°C) and the boiling point (155°C). At 2°C, the substance is liquid.

At 155°C, the substance changes from liquid to gas.

To calculate the heat gained for the change of 2°C liquid to 155°C liquid, specific heat capacity of the liquid (C) is needed.

To calculate the heat gained for the change of liquid to 155°C gas, heat of vaporization (D) is needed.

The specific heat of the solid is not used because the substance is changed from liquid to gas. it doesn't come in the state of solid.

Heat of fusion is not used, because it's used when there is a change from its state from a solid to a liquid,

The specific heat capacity of the gas is not used, because the substance only formes gas after reaching 155 °C

5 0

2 years ago