The structural form of the element Ge closely resembles the structure of C (diamond).
Diamond is composed of a lattice structure in which atoms of carbon are held together in a face centered cubic lattice.
Germanium, an element in the same group as carbon also forms a face centered cubic lattice that is very similar to that of diamond.
Hence, the structural form of the element Ge closely resembles the structure of C (diamond).
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Answer:
<h2>Dog's mitochondria lack the transport protein that transport pyruvate ( end product of glycolysis) across the outer mitochondrial membrane
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Explanation:
1. As given here that dog's mitochondria can use only fatty acids and also amino acids for their respiration, and as compared to others, Dong's cell produce more lactate then normal, this indicate that his mitochondrial membrane is different then others.
2. The aerobic phases of cellular respiration in eukaryotes occur within mitochondria. These aerobic phases are the TCA Cycle and the electron transport chain. Glycolysis occurs in the cytoplasm and the products of glycolysis enter into the mitochondria to continue cellular respiration.
3. These condition shows that dog's mitochondria lack the transport protein of mitochondria that moves pyruvate across the outer mitochondrial membrane.
Hi!
1) electrons
2) Chadwick
3) J. J. Thompson
4) Bohr
5) Rutherford
6) Dalton
7) Nucleus
I hope this helps!
The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>

The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
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