The 3 parts are
1) A five carbon ribose sugar
2) A Phosphate molecule
3) The four nitrogenous bases
I hope that's help !
Answer:
The rate of disappearance of 
for this period is
Explanation:
Initial concentration of = x = 0.0138 M
Final concentration of = y = 0.00886 M
Time elapsed during change in concentration = Δt = 374 s
Change in concentration ,![\Delta [NO_2]](https://tex.z-dn.net/?f=%5CDelta%20%5BNO_2%5D)
= y - x = 0.00886 - 0.0138 M = -0.00494 M
The rate of disappearance of for this period is:
![\frac{\Delta [NO_2]}{\Delta t}=\frac{-0.00494 M}{374 s}=-1.32\times 10^{-5} M/s](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B-0.00494%20M%7D%7B374%20s%7D%3D-1.32%5Ctimes%2010%5E%7B-5%7D%20M%2Fs)
Answer:1)1.99 M
Explanation:
Molarity is given as = moles solute/Liter solution
The solute which is LiOH is already given in moles as 1.495 moles
Given that solution is 750 mL, we convert to liters.
Liters of the solution= mL of the solution x (1 L/1000 mL)
750 mL x (1 L/1000 mL)
0.75 L
Molarity = moles solute/Liter solution
Molarity = 1.495 moles of LiOH/0.75 L of solution
Molarity = 1.99M
The molarity of this solution is 1.99M (moles per liter).
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</span><span>Cortisol
</span><span>Estrogen or testosterone
</span><span>Insulinlike growth factor-I (IGF-I)</span>
