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3 years ago

Explain the various modes of dispersal of seeds and fruits.

1 answer:

4 0

There are five main modes of seed dispersal: gravity, wind, ballistic, water, and by animals. Some plants are serotinous and only disperse their seeds in response to an environmental stimulus. Dispersal involves the letting go or detachment of a diaspore from the main parent plant.

Fruits and seeds dispersal is the process whereby fruits and seeds are scattered from their origin. The various ways by which fruit and seed are dispersed are known as agents of seed and fruit dispersal.

Check this link out for more information
https://qknowbooks.gitbooks.io/fruits-and-seeds/content/fruits_and_seeds_dispersal.html

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When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

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Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

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Density of solution = d = 1.00 g/mL

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$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

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$T_{initial}$ = initial temperature = $28.9^oC$

Now put all the given values in the above formula, we get:

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

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Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

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Read 2 more answers

Explain the various modes of dispersal of seeds and fruits.​

Explain the various modes of dispersal of seeds and fruits.