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Lemur [1.5K]
2 years ago
12

Help! Science Question! Brainliest can be yours!!!

Chemistry
2 answers:
Mkey [24]2 years ago
8 0

Answer:

CONSTANT

Explanation:

Carbon-14 is a radioactive isotope. It is found in the air in carbon dioxide molecules. The amount of carbon-14 in the air has stayed the same for thousands of years. There is a small amount of radioactive carbon-14 in all living organisms because it enters the food chain.

Once an organism dies, it stops taking in carbon-14. The carbon-14 it contained at the time of death decays over a long period of time, and the radioactivity of the material decreases.

evablogger [386]2 years ago
3 0

Answer:

constant

Explanation:

As long as an organism lives, the amount of carbon-14 it contains is <em>CONSTANT</em>.

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Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

<u>2) Entropy change for Decomposition of mercuric oxide </u>

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

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