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Masteriza [31]
2 years ago
10

How long is radioactive waste from nuclear plants radioactive?

Chemistry
2 answers:
Anettt [7]2 years ago
8 0

Answer:about 10,000 years b maybe

Explanation:Strontium-90 and cesium-137 have half-lives of about 30 years (half the radioactivity will decay in 30 years). Plutonium-239 has a half-life of 24,000 years. High-level wastes are hazardous because they produce fatal radiation doses during short periods of direct exposure.

Pavlova-9 [17]2 years ago
5 0
It can stay radioactive from 1,000 to 10,000 years. I honestly don’t know which answer to choose but it’s going to be between those two.
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Which has the highest density, water at 0c, water at 4c, water at 6c, water at 8c
Lelechka [254]

Answer:

water at 0C

Explanation:

The colder the water is, the denser it is, so the water here with the lowest temperature, is 0C

3 0
2 years ago
When the energy is removed what will happen to a gas?
creativ13 [48]

Answer:

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Explanation:

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8 0
3 years ago
Read 2 more answers
The normal boiling point of iodomethane, CH3I, is 42.43 8C, and its vapor pressure at 0.00 8C is 140. Torr. Calculate (a) the st
tigry1 [53]

Answer:

a=28600J; b=90.6 J/K; c=402 torr

Explanation:

(a) considering the data given

 Vapour pressure P1 =0  at Temperature T1 = 42.43˚C,

Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)

Using the Clausius-Clapeyron Equation

ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)

In 760/140 = ΔH/8.314 J/mol/K  × (1/315.58K -- 1/273.15K)

ΔH vap= +28.6 kJ/mol or 28600J

(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.

Since ΔG at boiling point is zero,

ΔS =(ΔH°vap/Τb)

 ΔS = 28600 J/315.58 K

= 90.6 J/K

(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)

ln P298 K/1 atm =  28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)

P298 K = 0.529 atm

                = 402 torr

8 0
3 years ago
What Group is Zn (zinc) in?<br> O A. 4<br> O B. 2A<br> O C. 12<br> O D. 30
Oxana [17]
I believe it’s (C.12)
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3 years ago
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