<u>Answer:</u> The molarity of calcium hydroxide in the solution is 0.1 M
<u>Explanation:</u>
To calculate the concentration of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is 
We are given:

Putting values in above equation, we get:

Hence, the molarity of
in the solution is 0.1 M.
<h3>Answer:</h3><h2>NADPH and ATP</h2><h3>Explanation:</h3>
When the light absorbs by the photosynthetic apparatus, the water molecules break down to produce essential material to continue the photosynthetic cycle as electron and H-ions. The H-ions becomes stored in thylakoid chamber and passed through the ATP synthetic machines to refill the ADP or AMP molecules. The generated electron passes through the various stages of photosynthetic series to finally produce the NADPH which also acts as energy transfer compound. At the PSI, the electron transfer to the NADP to generate NADPH.
Answer: The rates of production of
is
mol/Ls and
is 0.0195 mol/Ls.
Explanation:

Rate with respect to reactants is shown by negative sign as the reactants are decreasing with time and Rate with respect to products is shown by positive sign as the products are increasing with time.
Rate of the reaction=![-\frac{1}{4}\frac{[d[PH_3]}{dt}=\frac{[d[P_4]}{dt}=\frac{1}{6}\frac{[d[H_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B4%7D%5Cfrac%7B%5Bd%5BPH_3%5D%7D%7Bdt%7D%3D%5Cfrac%7B%5Bd%5BP_4%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7B%5Bd%5BH_2%5D%7D%7Bdt%7D)
Rate of decomposition of 
Rate of production of 
Rate of production of 