Answer:
7 orbitals
Explanation:
An f sublevel has 7 orbitals
Answer: A material that readily transmits energy is a conductor, while one that resists energy transfer is called an insulator .
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Answer:
Role is defined below
Explanation:
A small GTP-binding protein, is an important module of the signal transduction pathway used by growth factors to initiate cell growth and differentiation. Cellular activation with growth factors such as epidermal growth factor (EGF) induces Ras to move from an inactive state linked to GDP to an active state linked to GTP. In recent times, a mixture of genetic and biochemical studies has resulted in the elucidation of a signaling pathway that leads from growth factor receptors to Ras. After joining EGF, the EGF receptor tyrosine kinase is activated, which leads to receptor auto phosphorylation in multiple tyrosine residues. Signaling proteins with homology domains Src 2 (SH2) then bind to these phosphorylated residues in tyrosine, initiating multiple signaling cascades. Distinct of these SH2 area proteins, Grb2, exists in the cytoplasm in a preformed complex with a second protein, Son of Sevenless (Sos), which can catalyze the Ras GTP / GDP exchange. After stimulation of the growth factor, the phosphorylated EGF receptor with tyrosine binds to the Grb2 / Sos complex and translocates it to the plasma membrane. It is believed that this translocation brings Sos closer to Ras, which leads to the activation of Ras. In dissimilarity, the insulin receptor does not bind Grb2 directly, but rather induces the tyrosine phosphorylation of two proteins, the substrate-1 insulin receptor and Shc, which bind to the Grb2 / Sos complex. Once Ras is activated, a cascade of protein kinases that are important in a myriad of growth factor responses is stimulated.
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
