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Vitek1552 [10]
3 years ago
15

What are parts of the nervous system

Chemistry
1 answer:
Vlad1618 [11]3 years ago
6 0

Answer:

the brain and the spinal cord are the two main parts.

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How many parts does CsF ( Cesium Fluoride ) have?
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Review the PCR (Calvin-Benson-Basham) cycle and the photochemical (light) reactions. If photosynthesizing green algae are provid
stepladder [879]

Answer:

Explanation:

THE PHOTOCHEMICAL(LIGHT) REACTIONS :This is a phase of photosynthesis where sunlight is used as a source of energy to manufacture two chemical compounds which are "Reduced nicotinamide adenine dinucleotide phosphate"-NADPH and "Adenosine triphosphate"-ATP.This phase of photosynthesis involves 4 steps or reactions which are :

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STEP 2 : PHOTOLYSIS OF WATER - Here the energy absorbed by the chlorophyll molecules are used to split water molecules into H+ ions and OH-- ions.

STEP 3:Formation of NADPH -The hydrogen ions (H+) produced reacts with an NADP ( an electron carrier in the chlorophyll) to form NADPH.

STEP 4: FORMATION OF ATP - The high energy generated from the electron transfer process or chain is used to add a phosphate group to ADP (Adenosine dphosphate) to form ATP.

DARK PHASE :In this phase of photosynthesis,the NADPH generated in the light phase is used as a reducing equivalent to reduce CO2 to form Glucose (food) using the ATP generated as a source of energy.

7 0
3 years ago
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
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