This is a kind of an asexual reproduction called binary fission. This involves division of the parent cell into two identical cells. This is common for prokaryotic reproduction like archaea and bacteria and it also happens in some single-celled eukaryotes such as protists and unicellular fungi.
I think it would be ionization
Answer: The ground-state electronic configuration of
1) ![Ru^{2+}=[Kr]4d^6](https://tex.z-dn.net/?f=Ru%5E%7B2%2B%7D%3D%5BKr%5D4d%5E6)
2) ![W^{3+]=4f^{14}5d^3](https://tex.z-dn.net/?f=W%5E%7B3%2B%5D%3D4f%5E%7B14%7D5d%5E3)
Explanation: The electronic configuration of elements is defined as the distribution of electrons around the nucleus of that element. It depends on the atomic number of the element.
Atomic number of the element = Number of electrons.
Atomic number = 44
Number of electrons = 44
Electronic configuration of Ru-element = ![[Kr]4d^75s^1](https://tex.z-dn.net/?f=%5BKr%5D4d%5E75s%5E1)
To form
, 2 electrons are released from the neutral Ru-element.
So, the electronic configuration of ![Ru^{2+}=[Kr]4d^6](https://tex.z-dn.net/?f=Ru%5E%7B2%2B%7D%3D%5BKr%5D4d%5E6)
Atomic number = 74
Number of electrons = 74
Electronic configuration of W-element = ![[Xe]4f^{14}5d^46s^2](https://tex.z-dn.net/?f=%5BXe%5D4f%5E%7B14%7D5d%5E46s%5E2)
To form
, 3 electrons are released from the neutral W-element.
So, the electronic configuration of ![W^{3+}=[Xe]4f^{14}5d^3](https://tex.z-dn.net/?f=W%5E%7B3%2B%7D%3D%5BXe%5D4f%5E%7B14%7D5d%5E3)
Answer:
Types of atomic orbitals present in the third principal energy are <u>s, p and d only .</u>
Explanation:
- <u>OPTION A-: s and p atomic orbitals -</u> these two orbitals are present in second principal energy level. Therefore , the option is incorrect.
- <u> OPTION B-: p and d only -</u> This option is wrong as there is no such principal level energy where , s atomic orbital is absent .
- <u>OPTION C-: s , p and d only -</u>these orbitals are present in<u> third principal energy level</u>. The third major level of energy has one orbital, three orbitals of p, and five orbitals of d, each of which can contain up to 10 electrons. The third stage thus holds a maximum of 18 electrons. This option is correct .
- <u>OPTION D-: s , p, d and f only -</u>There is also a f sublevel at the <u>fourth and higher stages,</u> containing seven f orbitals, which can accommodate up to 14 electrons at most. Therefore, up to 32 electrons will hold the fourth level: 2 in the s orbital, 6 in the three p orbitals, 10 in the five d orbitals, and 14 in the seven f orbitals. This option is incorrect .
<u>Thus , the correct option is C (s , p and d only .)</u>
<u>Answer:</u> The Gibbs free energy of the reaction is 21.32 kJ/mol
<u>Explanation:</u>
The chemical equation follows:

The equation used to Gibbs free energy of the reaction follows:

where,
= free energy of the reaction
= standard Gibbs free energy = 29.7 kJ/mol = 29700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314J/K mol
T = Temperature = ![37^oC=[273+37]K=310K](https://tex.z-dn.net/?f=37%5EoC%3D%5B273%2B37%5DK%3D310K)
= Ratio of concentration of products and reactants = ![\frac{\text{[Oxaloacetate]}[NADH]}{\text{[Malate]}[NAD^+]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%5BOxaloacetate%5D%7D%5BNADH%5D%7D%7B%5Ctext%7B%5BMalate%5D%7D%5BNAD%5E%2B%5D%7D)
![\text{[Oxaloacetate]}=0.130mM](https://tex.z-dn.net/?f=%5Ctext%7B%5BOxaloacetate%5D%7D%3D0.130mM)
![[NADH]=2.0\times 10^2mM](https://tex.z-dn.net/?f=%5BNADH%5D%3D2.0%5Ctimes%2010%5E2mM)
![\text{[Malate]}=1.37mM](https://tex.z-dn.net/?f=%5Ctext%7B%5BMalate%5D%7D%3D1.37mM)
![[NAD^+]=490mM](https://tex.z-dn.net/?f=%5BNAD%5E%2B%5D%3D490mM)
Putting values in above expression, we get:

Hence, the Gibbs free energy of the reaction is 21.32 kJ/mol