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ANEK [815]
3 years ago
8

Some radioactive nuclides have very short half-lives, for example, I-31 has a half-life of approximately 8 days. Pu-234, by comp

arison has a half-life of 24,000 years. Explain why both of these examples are dangerous, even though their half-lives are very different. Be sure to describe the different major types of radiation, and their hazards. (Radioactive Decay and Half-Life)
Chemistry
1 answer:
lorasvet [3.4K]3 years ago
6 0

Answer:

Here's what I find.

Explanation:

Iodine-131

Iodine-131 is both a beta emitter and a gamma emitter.

_{53}^{131}\text{I}\longrightarrow \, _{54}^{131}\text{Xe} +\, _{-1}^{0}\text{e} +\, _{0}^{0}\gamma

About 90 % of the energy is β-radiation and 10 % is γ-radiation. Both forms are highly energetic.

The main danger is from ingestion. The iodine concentrates in thyroid gland, where the β-radiation destroys cells up to 2 mm from the tissues that absorbed it.

Both the β- and γ-radiation cause cell mutations that can later become cancerous. Small doses, such as those absorbed from the nuclear disasters in the Ukraine and Japan, can cause cancers years after the original iodine has disappeared.

Plutonium-239

Plutonium-239 is an alpha emitter.

_{94}^{239}\text{U} \longrightarrow \, _{92}^{235}\text{Xe} + \, _{2}^{4}\text{He}

Alpha particles cannot penetrate the skin, so external exposure isn't much of a health risk.

However, they are extremely dangerous when they are inhaled and get inside cells. They travel first to the blood or lymph system and later to the bone marrow and liver, where they cause up to 1000 times more chromosomal damage than beta or gamma rays.

It takes about 20 years for plutonium to be eliminated from the liver around 50 years for from the skeleton, so it has a long time to cause damage.

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-Calculate the number of meters traveled by a girl who swims at a speed of 1.7
Dafna11 [192]

Answer:

Explanation:

Use equation:

1.7 m       59 sec

-------  X  -----------  = 100.3 m

sec              1

The Seconds on the top and bottom cancel each other out so the unit of measurement left is Meters.

3 0
3 years ago
How much heat is absorbed when 90.5 g of ice is heated from -11.0 °C to 145.0 °C?
Nadusha1986 [10]

Answer:

Q(total) = 283Kj

Explanation:

5 Heat Transitions …

Specific Heats => c(s) = 0.50cal/g∙⁰C,  c(l) = 1.0 cal/g∙⁰C, c(g) = 0.48 cal/g∙⁰C

Phase Transition Constants => ΔHᵪ = Heat of Fusion = 80 cal/g; ΔHᵥ = Heat of Vaporization = 540cal/g

Note => Phase change regions => no temp. change occurs when 2 phases are in contact (melting and evaporation). Only when single phase substance exists (s, l or g) does temperature change occur. See heating curve for water diagram. The increasing slopes are temperature change regions and heat flow is given by Q =mcΔT. The horizontal slopes are phase changes ( melting & evaporation) and heat flow for each of those regions is given by Q = m·ΔH. Each transition energy is calculated individually (see below) and added to obtain the total heat flow needed.

Q = mcΔT for temperature change regions of the heating curve (single phase only)

Q = m∙ΔH for phase transition regions of the heating curve (2 phases in contact)

Solid (ice) => Melting Pt  => Q(s) = mcΔT = (90.5g)(0.50cal/g∙⁰C)(11⁰C) = 478 cal

Melting (s/l) => Liquid (water) =>   Q(s/l) = m∙ΔHᵪ = (90.5g)(80cal/g) = 7240 cal

Liquid (water) => Boiling Pt => Q(l) = mcΔT = (90.5g)(1.0cal/g∙⁰C)(100⁰C) = 9050 cal

Boiling (l/g) => Gas (steam) => Q(l/g) = m∙ΔHᵥ = (90.5g)(540cal/g) = 48,870 cal

Gas (steam) => Steam @ 145⁰C => Q(g = mcΔT = (90.5g)(0.48cal/g∙⁰C)(45⁰C) = 2036 cal

Total Heat Transfer (Qᵤ) = Q(s) + Q(s/l) + Q(l) + Q(l/g) + Q(g)  

                                 = 478cal +7240cal + 9050 cal + 48,870cal + 2036cal

                                 = 67,674 cal x 4.184 j/cal = 283,148 joules = 283 Kj

4 0
4 years ago
Hydrogen ion reacts with zinc to produce_____ gas?
WINSTONCH [101]

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

Hydrogen ion reacts with zinc to produce\:\pmb{\underline{\red{\sf{Zinc \:hydride    }}}}\:gas

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

6 0
3 years ago
The group of cells that are organized into columns and rows
Brrunno [24]

Explanation:

vertical columns and horizontal rows, hope it helps

5 0
3 years ago
Consider the reaction: CO2(g) + CCl4(g) ⇌ 2 COCl2(g) ΔG° = 46.9 kJ Under the following conditions at 25 oC: LaTeX: P_{CO_2}P C O
Mamont248 [21]

Answer : The value of \Delta G_{rxn} is, -47.0kJ/mole

Explanation :

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln K_p   ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction

\Delta G_^o =  standard Gibbs free energy  = 46.9 kJ

R = gas constant = 8.314 J/mole.K

T = temperature = 25^oC=273+25=298K

K_p = equilibrium constant

First we have to calculate the value of K_p.

The given balanced chemical reaction is,

CO_2(g)+CCl_4(g)\rightarrow 2COCl_2(g)

The expression for reaction quotient will be :

K_p=\frac{(p_{COCl_2})^2}{(p_{CO_2})\times (p_{CCl_4})}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

K_p=\frac{(0.653)^2}{(0.459)\times (0.984)}

K_p=0.944

Now we have to calculate the value of \Delta G_{rxn} by using relation (1).

\Delta G_{rxn}=\Delta G^o+RT\ln K_p

Now put all the given values in this formula, we get:

\Delta G_{rxn}=-46.9kJ/mol+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (0.944)

\Delta G_{rxn}=-47.0kJ/mol

Therefore, the value of \Delta G_{rxn} is, -47.0kJ/mole

3 0
3 years ago
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