Answer:
2 hydrogen and 1 oxygen atoms
Explanation:
Answer:
333.3mL
Explanation:
Using the formula as follows:
C1V1 = C2V2
Where;
C1 = initial concentration (M)
C2 = final concentration (M)
V1 = initial volume (mL)
V2 = final volume (mL)
According to the information provided in this question,
C1 = 4.00M
C2 = 1.50M
V1 = 125mL
V2 = ?
Using C1V1 = C2V2
4 × 125 = 1.5 × V2
500 = 1.5V2
V2 = 500/1.5
V2 = 333.3mL
Therefore, the CuSO4 solution needs to be diluted to 333.3mL to make 1.50 M solution.
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.
Properties change within a Row (or period) as the number of valence electrons in the outer shell increase
Answer : The concentration of 
and 
at equilibrium is, 0.0158 M and 0.00302 M respectively.
Explanation :
First we have to calculate the concentration of 
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
Initial conc. 0.0163 0.00415 0.00276
At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)
As we are given:
Concentration of 
at equilibrium = 0.00467 M
That means,
(0.00415+x) = 0.00467
x = 0.00026 M
Concentration of at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M
Concentration of at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M
