What is the difference between acute and chronic radiation poisoning

1 answer:

4 0

Acute: Can be repaired by the body using white blood cells.

Chronic: High doses causing internal bleeding and breakdown of cells causing slow death, 90% of the times cannot be fixed by any medicines.

You might be interested in

43 Which element reacts spontaneously with 1.0M HCl (aq) at room temperature?

KonstantinChe [14]

The correct answer is (4) zinc~

According to the reactivity series of metals, only zinc react with HCl (aq) at room temperature.

Hope this helps~

4 0

3 years ago

As stated in the article, “As Sticky as a Gecko . . . but Ten Times Stronger!,” the adhesive the researchers developed sticks be

alexandr1967 [171]

Answer:

shear adhesion

Explanation:

edge

6 0

3 years ago

Read 2 more answers

Why is the moon unique

Ainat [17]

Answer:

Explanation:

well there is many things in our solar system and a moon is one of them. the moons in our solar system are big and small and orbit our planets. the moon is unique because it shines in the dark while it lets the sun shine in the light.

Thats what i would say but its an eh answer

8 0

2 years ago

Read 2 more answers

Example 13:0.29 grams of a hydrocarbon with vapour density 29 when burnt completely in oxygen produce 448 ml of carbon dioxide a

Juliette [100K]

Answer:

Given, 0.29 g of hydrocarbon produces 448ml of CO2 at STP. then, C2H5 is the emperical formula of hydrocarbon . n = 2 , hence, molecular formula will be C4H10

6 0

2 years ago

A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so

omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: $AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s)$ .

Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
Secondly, addition of liquid ammonia would form a precipitate with silver: $AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)$ ; Silver hydroxide at higher temperatures decomposes into black silver oxide: $2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)$ .
Thirdly, we also know we have $Hg_2^{2+}$ in the mixture, since addition of potassium chromate produces a yellow precipitate: $Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s)$ . The latter precipitate is yellow.

3 0

3 years ago