2 NH4+ contribute 8 mols of H
There is 1 more mol of H infront of the PO4+3
9 mols H <<<<<===== answer.
Answer : The correct option is 'X' (mitochondria).
Explanation :
In the given animal cell diagram,
W represent the vacuole.
X represent the mitochondria.
Y represent the nucleolus.
Z represent the cytoplasm.
The function of vacuole, it stores waste products and balance the pH level of the cell.
The function of mitochondria, most of the respiration reaction occurs in which carbon dioxide is produced by utilizing oxygen.
The function of nucleolus, it contains the hereditary information.
The function of cytoplasm, most of the chemical reaction occurs.
Therefore, the correct answer is 'X' (mitochondria).
Answer:
well what's the question? we can't see it
Answer:
See Explanation
Explanation:
The colour of many transition metal complexes stem from transitions of electrons between energy levels. These transitions are governed by the spin selection rules and the colour is determined by the magnitude of crystal field splitting.
According to the spin selection rules, transitions in which ΔS = 0 are forbidden. Hence, a Mn^2+high spin compound is expected to be colourless. However, contrary to the spin selection rules Mn^2+high spin compounds do exhibit transitions in which the intensity is only about one-hundredth of the intensity of the spin allowed transitions. Thus many Mn^2+ high spin compounds such as Mn(NO3)2 are very pale pink or off white.
Note also that the crystal field stabilization energy of Mn^2+ which is a d^5 low spin ion is zero hence the very pale colour observed.
K4[Mn(CN)6] is deep blue as a result of charge transfer. Also, the compound exhibits an observed crystal field stabilization energy because it is a d^5 low spin compound hence the observed colour. Its low spin nature is because the cyanide ion is a strong field ligand hence it causes a greater magnitude of crystal filed splitting.
The following compounds are colourless;
Zn(NO3)2
CdSO4
AgClO3
One thing that is common to all the compounds listed above is that they are all d^10 compounds. This means that they all possess completely filled d-orbitals hence they are colourless.
Answer:
Atomic radius of Strontium is 27.38pm
Explanation:
In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:
a² + a² = b² = (4r)²
2a² = 16r²
a = √8 r
As edge length of Strontium is 77.43pm:
77.43pm / √8 = r
27.38pm = r
<h3>Atomic radius of Strontium is 27.38pm</h3>