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a_sh-v [17]
3 years ago
15

How can radiation be life-threatening and at the same time be useful in saving lives?

Chemistry
1 answer:
jek_recluse [69]3 years ago
5 0

Answer:

Explanation:

Radiation can either kill cells or damage the DNA within them, which damages their ability to reproduce and can eventually lead to cancer. However, it may be useful because it minimi damage to DNA

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What is the significance sodium hydroxide and hydrochloric acid
hammer [34]

Answer:

They are a base and an acid,  so they neutralize each other.

Explanation:

Sodium hydroxide is a strong base. Hydrochloric acid is a strong acid.

They react with (neutralize) each other to form a salt (sodium chloride) and water.

NaOH + HCl ⟶ NaCl + H₂O

5 0
3 years ago
HELP!!!!!!!!!!! *What are some of the safety concerns for Barium sulfate & Barium Sulfide? Explain
Alex

BaSO₄ is relatively harmless, but BaS is highly toxic.

BaSO₄ is quite insoluble (240 µg/100 mL). It is a <em>mild irritant</em> in cases of skin contact and inhalation. However, it is <em>safe enough</em> that health professionals ask patients to drink a suspension of BaSO₄. The Ba is opaque to X-rays, so it makes the stomach and intestines more visible to radiographers.

BaS is soluble (7.7 g/100 mL). It reacts slowly with water and more rapidly in the acid conditions of the stomach to <em>release H₂S</em>.

BaS + 2HCl ⟶ BaCl₂ + H₂S

An H₂S concentration of 60 mg/100 mL can be <em>fatal within 30 min</em>.

<em>Don’t eat barium sulfide!</em>

3 0
3 years ago
Which of the following is considered a form of technology?
anygoal [31]

Answer:

The answer is B. Hope this helps! please give me brainliest!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! =)

8 0
3 years ago
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!!!!!!!!!!!!SCIENCE!!!!!!!!!!!!<br><br> How are radio waves used on Earth?
nika2105 [10]

Answer:

artificially generated radio waves are used for fixed and Mobile Radio communication broadcasting radar and other navigation systems communication satellites computer networks and innumerable other applications

7 0
3 years ago
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A 45.0 g sample of a metal at 85.6 °C is placed in 150.0 g of water at 24.6 °C. The final temperature of the system is 28.3 °
earnstyle [38]

Answer:

904.014 j/kgk

Explanation:

Mass of metal = 45g

Temperature of metal = 85.6°

Mass of water = 150

Temperature of water = 24.6

Final temperature of system = 28.3

Heat lost by metal = Heat gained by water

m1 * c1 * dt = m2 * c2 * dt

Q = quantity of heat

Q = m*c*dt

dt = change in temperature

dt of water = 28.3 - 24.6 = 3.7

dt of metal = 85.6 - 28.3 = 57.3

Specific heat capacity of water, c = 4200

(45 * 10^-3) * c * 57.3 = (150 * 10^-3) * 4200 * 3.7

2.5785c1 = 2331

c1 = 2331 / 2.5785

= 904.01396

= 904.014 j/kgk

3 0
3 years ago
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