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alexgriva [62]
2 years ago
7

Horticulturists know that, for many plants, dark green leaves are associated with low light levels and pale green with high leve

ls. (a) Use the photon theory to explain this behavior.
Chemistry
1 answer:
nikitadnepr [17]2 years ago
7 0
  • In the reduction process of photosynthesis, electrons are transported from water to carbon dioxide.
  • According to photon theory here the photons behave as light energy particle.
  • By capturing sun energy (photons), chlorophyll aids in this process.An electron in the chlorophyll molecule gets excited from a lower to a higher energy state when chlorophyll receives energy from sunlight.
  • Excessive sunshine is the main cause of leaves turning a darker green color. Lack of sunlight causes a plant's leaves to create more chlorophyll, which is what gives the plant its green hue.
  • The color of the leaf is darker green than typical because of the increase in chlorophyll concentration per unit area.
<h3>What is photon theory?</h3>

The photon theory of light states that photons have simultaneous particle and wave behavior. pass through empty space at a constant speed, or "the speed of light". possess rest energy and zero mass.

Learn more about photon theory here:

brainly.com/question/3479307

#SPJ4

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Write net ionic equation to show the reaction of aqueous hg2(no3)2 with aqueous sodium chloride to form solid hg2cl2 and aqueous
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The net ionic equations in the creation of solid Hg2Cl2 and aqueous Sodium Nitrate (NaNO3) would be:

Hg2(NO3)2 + 2 NaCl --------> Hg2Cl2 + 2 NaNO3 <span>
Hg2+2 + 2NO3-1 + 2Na+ Cl- ----> Hg2Cl2 + 2Na+ + 2NO3-1 
<span>Hg2+2 + 2Cl-1 ------------> Hg2Cl2</span></span>

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Dekameter

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The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g)
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Answer:

\Delta H=-29.7kJ

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

N_2+3H_2\rightarrow 2NH_3

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

n_{NH_3}^{by\ H_2}=1.96gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.647molNH_3\\\\  n_{NH_3}^{by\ N_2}=9.51gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.679molNH_3

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

\Delta H=0.647molNH_3*\frac{-45.9kJ}{1molNH_3}\\\\ \Delta H=-29.7kJ

Best regards!

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3 years ago
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The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.

The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).

The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).  

The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.

Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.    


4 0
3 years ago
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