All categories

2 years ago

Explain why spent nuclear fuel rods have to be treated as hazardous waste.

2 answers:

4 0

If the nuclear fuel rods are still dripping with nuclear fuel or if its as clean as a baby, it doesn't matter since it can still be harmful to living things in a microscopic form, nothing in the world is 100% clean, even after you take a shower your body isn't even 100% clean, your body is crawling with germs

mojhsa [17]2 years ago

3 0

Technically it's not such a waste anymore but it is dangerous chemicals that could kill others around it.

Let me know if you need additional help in private message.

-GeVontae The Brainly team.

You might be interested in

Plants are considered

wariber [46]

Because they create their own food, they are considered as producers.

6 0

2 years ago

What domain is an onion

Nana76 [90]

The domain that is an onion is the Eukaryote.

3 0

3 years ago

Read 2 more answers

What is a logical way that you could begin to investigate why your tire had low air pressure? Look to see if there is a hole in

MrRissso [65]

Answer:

Any of the answers given will work

Explanation:

I literally just did it.

7 0

3 years ago

Read 2 more answers

SOME ONE HEEEEELLLLPPPPPPPP PLEAAASEEE

stellarik [79]

Answer:

look up chemical board it will help you

Explanation:

6 0

2 years ago

what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C

svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

$\frac{V}{T}=k$

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

$\frac{V1}{T1}=\frac{V2}{T2}$

In this case:

V1= 0.730 L
T1= 28 °C= 301 °K (0°C= 273°K)
V2= ?
T2= 0°C= 273 °K

Replacing:

$\frac{0.730 L}{301K}=\frac{V2}{273K}$

Solving:

$V2=273K*\frac{0.730L}{301K}$

V2=0.662 L

The volume of the air is 0.662 L

6 0

3 years ago