All categories

2 years ago

The primary toxicity of pulmonary agents is to the and follows __ of smoke, particles, vapors, or gases

2 answers:

6 0

The words that best complete the given statement above would be "airway" and "inhalation." Therefore, the complete statement would be "<span>The primary toxicity of pulmonary agents is to the airway and follows inhilation of smoke, particles, vapours, or gases." These air pollutant would likely cause respiratory problems once inhaled in excessive amounts.</span>

Sphinxa [80]2 years ago

4 0

Answer:

The primary toxicity of pulmonary agents is to the airway and follows inhalation of smoke, particles, vapors, or gases

Explanation:

Toxic substances which are in the form of gases, dust, vapors, fumes, fibers, etc. are able to penetrate organisms by inhalation through both the upper airways and the alveoli.

The respiratory system consists of the upper airways and the lungs.

The pathway of toxic substances through the respiratory system will depend on the body's reaction to these agents. Toxic substances are often grouped into particles and, depending on their size, do not even reach all the respiratory compartments.

You might be interested in

Which best describes the bond that would form between lithium (Li) and fluorine (F)?

Vladimir79 [104]

They form an ionic bond by exchanging one electron

3 0

2 years ago

Read 2 more answers

If we consider the challenges posed by the Great Plate Count Anomaly, what is the primary advantage of using culture independent

nordsb [41]

Culture independent techniques is simply the use of methods in microbiology that are not based on or relies on cultivation inorder to study a microorganism in a defined ecosystem.

Independent based techniques offer a variety of advantages over the culture dependent methods.

The primary advantage of using culture independent techniques in microbiology is that culture independent techniques has the ability to identify or point out a large proportion of the microbial diversity readily and that identification on a normal state will be difficult to be observed using culture based studies or culture independent techniques.

8 0

2 years ago

Can we still recover the lost biodiversity? Why or why not?

photoshop1234 [79]

Yes, we can recover it. But not the same as before. Maybe we will make a new one. By planting trees and cleaning the area.The loss of biodiversity has two significant impacts on human health and the spread of disease. First, it increases the number of disease carrying animals in local populations. ... At the same time, habitat fragmentation brings humans in closer and more frequent contact with these disease carrying species.Participating in Biodiversity Conservation. Identify locations of critical wildlife habitat for species at risk and the threats to these areas. ... Keep vehicles on main roads to reduce the spread of weeds and disturbance to wildlife. Monitor and assess your pets' impact on biodiversity.

Hope this helped <3

8 0

2 years ago

The state of health and functioning of the liver is often assessed with dye-tracer techniques. The dye used most frequently is b

elena-s [515]

Answer

1.Radioactive or chemical decay

2. X(t) = Xoe^-kt

lnY(t) = -t + lnYo

4. 6.07mg

Explanation:

Let the liver and and blood compartment be represented by the symbol L and B respectively

For the liver

Suppose a first Order removal process started with an amount X, in which amount b disappeared in time t, the process is decay process which can be represented as follows,

∫dX/dt = -K1X

By rearrangement and integration;

∫dX/X = -K1t

ln X = -K1t + C

Since At t= 0, X = Xo then

C = lnXo

The equation becomes:

lnX = -K1t + lnXo

lnX - lnXo = -K1t

ln(X/Xo) = -K1t

X/Xo = e^-kt

X(t) = Xoe^-kt

X(t) = Xoe^-0.5t........(2)

Similarly for B (blood), suppose a first order flow flow of the dye move from the blood to the liver, let Y be the initial concentration, and amount b that has flown to the liver in time t

B---------> B(t)

t=0 Yo 0

t=t. Y-b. b

dY/dt = -K12(Y-b)...........(3)

Let Y-b = Y(t)

∫dY/dt = -∫K12t

By rearrangement and integration;

∫dy/Y(t) = ∫-K12dt

lnY(t) = -K12t + C1

at t= 0, C1 = ln(Yo)

Therefore ln Y(t) = -K12t + lnYo

But K12 =1

ln Y(t) = -t + lnYo.............(4)

(3) The assumptions used here

is that of a decay for the liver . The amount remaining taking as the amount of a substance

(4) using the equation 2,

X(t)= Xo e^-K1t........(2)

For time t = 1hour, and an initial amount X = 10mg, K = 0.5

X(t) = 10× e^-0.5

X(t) = 10 × 0.607

X(t) = 6.07mg

(5) within the scope of information presented, I have no data to make this judgment.

3 0

2 years ago

11. Using the diagram to the left,if line n bisects QR find QP

trapecia [35]

Answer:

QP = 41

Explanation:

Its geometry its easy...

Since its getting cut through the middle that means they equal each other

3x + 5 = 5x - 19 (Use a calculator)

x = 12

Since your solving to find QP you plug it into QP

3(12) + 5 = 41

4 0

2 years ago

The primary toxicity of pulmonary agents is to the ________ and follows __________ of smoke, particles, vapors, or gases

The primary toxicity of pulmonary agents is to the and follows __ of smoke, particles, vapors, or gases