10. What is the mass of 2.6 moles of Ag?

Explain the various modes of dispersal of seeds and fruits.

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

4 0

3 years ago

Summarize tornado in 6 words:

guajiro [1.7K]

Tornado is a spiral of debris

7 0

2 years ago

Read 2 more answers

How does ph affect photosynthesis?

liraira [26]

Photosynthesis is the process by which plants create their food. It involves the operation of enzymes in the plant cells , and they work best at certainpH levels. ... Thus, as the plant's pH drifts away from the best pH, the rate of photosynthesis will decrease.

4 0

3 years ago

1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en

Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

$n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles$

$n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles$

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

$n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles$