All categories

2 years ago

Help plz:))) I’ll mark u Brainliest

Chemistry

1 answer:

loris [4]2 years ago

3 0

Answer:

We report an unusual case of mercury vapor poisoning from using a heated tobacco product. The suspect had added grains of mercury into 20 cigarettes in a pack. When a 36-year-old Japanese man inserted one of these cigarettes into the battery powered holder, it was heated to a temperature of 350 °C, and he inhaled vaporized mercury. After using 14 of the cigarettes over 16 h, he noticed he had flu-like symptoms so he visited the hospital. Although no physical abnormalities were revealed, 99 μg/L of mercury was detected in his serum sample. His general condition improved gradually and his whole blood mercury level had decreased to 38 μg/L 5 days later. When the remaining six cigarettes in the pack were examined, many metallic grains weighing a total of 1.57 g were observed. Energy dispersive X-ray fluorescence spectrometry confirmed the grains as elemental mercury. Accordingly, the victim was diagnosed with mercury poisoning. Because the mercury was incorporated into cigarettes, an unusual and novel intoxication occurred through the heating of the tobacco product. Both medical and forensic scientific examination confirmed this event as attempted murder.

Explanation:

You might be interested in

Si una masa dada de hidrógeno ocupa 40 litros a 700 grados torr. ¿Qué volumen ocupará a 1 atmósfera de presión? (dar la presión

Svetach [21]

Answer:

$V_2=36.84L$

Explanation:

Hola,

En este caso, podemos usar la ley de Boyle, la cual nos permite analizar el comportamiento volumen-presión en un gas ideal de manera inversamente propocional:

$P_1V_1=P_2V_2$

Así, dado el volumen y la presión inicial, la cual se convierte a atmósferas (760 torr = 1atm), calculamos el volumen final a 1 atm como se muestra a continuación:

$V_2=\frac{P_1V_1}{P_2}=\frac{700torr*\frac{1atm}{760torr}*40l }{1atm}\\ \\V_2=36.84L$

Saludos!

7 0

3 years ago

Read 2 more answers

Particles of light are known as

Mashcka [7]

HEY DEAR..

The particles of light known as photon.

HOPE ITS HELPFULL

6 0

2 years ago

Read 2 more answers

The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$