Answer: Static electricity will help in seed dispersal in cone bearing seeds.
Explanation:
Static electricity is the stationary electric charge produced by the friction. The static electricity can be produced by the turbulence or air currents in the storm clouds. This is responsible for the cause of lightening phenomena.
The cone bearing seeds exhibit cones that are hard and woody in nature. The static electricity due to the effect of lightening generate electrical shock which breaks open the cone and disperse the seeds. Thus the static electricity is important for seed dispersal in cone bearing trees.
The question is asking to state the substance that is acting as the Bronsted Lowry base in the forward reaction, base on my research, I would say that the substance acting to said formulas is the NH3. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification
3 molecules of O2 are used in this reaction.
Hope this helps!
That means that one mole of sulfur has a mass of 32.065 grams. b) This compound contains two hydrogen atoms and one oxygen atom. To find the molar mass of H2O, we need to add the mass of two hydrogen atoms plus the mass of one oxygen atom. We get: 2(1.008)+16.00=18.016g/mol.
In order to find the two statements, we must first define what the enthalpy of formation and the enthalpy of reaction mean.
Enthalpy of formation:
The change in enthalpy when one mole of substance is formed from its constituent elemetns at standard state.
Enthalpy of reaction:
The change in enthalpy when a reaction occurs and the reactants and products are in their standard states.
Now, we check the statements. The true ones are:
The Hrxn for C(s) + O₂(g) → CO₂(g) is the same as Hf for CO₂
This is true because the formation of carbon dioxide requires carbon and oxygen in their standard states.
The Hf for Br₂<span>(l) is 0 kJ/mol by definition.
Because the bromine is present in its standard state, the enthalpy of formation is 0.
</span><span>The Hrxn for the reaction 1.5H</span>₂<span>(g) + 0.5N</span>₂<span>(g) </span>→ <span>NH</span>₃<span>(g) is the same as the Hf for NH</span>₃<span>(g)
The reactants and products are present in their standard state, and the reaction is the same as the one occurring during the formation of ammonia.
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