lead is a toxic metal that affects the central nervous system. a pbpb-contaminated water sample contains 0.0013 %pb%pb by mass.

Chemistry

1 answer:

Rus_ich [418]1 year ago

7 0

0.0013% = 0.0013 g/100 ml water, or 1.3 mg/100 ml water

so for 145 mg, you need 145×100/1.3 = 11154 ml water or 11.154 L of water. Lead is a toxic metal that affects the central nervous system.

Extensively talking, the frightened gadget is organized into essential parts, the central nervous system(CNS) and the peripheral nervous system (PNS). The CNS is the processing center of the frame and includes the mind and the spinal cord. Both of these are protected through 3 layers of membranes referred to as meninges.

The principal anxious gadget is the frame's processing center. The brain controls most of the capabilities of the frame, inclusive of cognizance, movement, thinking, speech, and the five senses of seeing, listening to, feeling, tasting and smelling. The spinal twine is an extension of the mind.

The brain is what controls all the frame's features. The spinal twine runs from the brain down via the again. It carries threadlike nerves that department out to each organ and frame part. This network of nerves relays messages from side to side from the mind to distinctive parts of the body.

Learn more about central nervous system here:- brainly.com/question/2114466

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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:

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