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kirza4 [7]
2 years ago
11

Radioactive cobalt-60 is frequently used in treating cancer. It took 24 years for a 10 gram sample to decay to 0.625 grams. What

is the half-life of the sample?
Chemistry
1 answer:
-BARSIC- [3]2 years ago
5 0

Answer:

Half life is 6 years.

Explanation:

T½ = In2 / λ

Where λ = decay constant.

But N = No * e^-λt

Where N = final mass after a certain period of time

No = initial mass

T = time

N = 0.625g

No = 10g

t = 24 years

N = No* e^-λt

N / No = e^-λt

λ = -( 1 / t) In N / No (inverse of e is In. Check logarithmic rules)

λ = -(1 / 24) * In (0.625/10)

λ = -0.04167 * In(0.0625)

λ = -0.04167 * (-2.77)

λ = 0.1154

T½ = In2 / λ

T½ = 0.693 / 0.1154

T½ = 6.00 years.

The half life of radioactive cobalt-60 is 6 years

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Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2
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Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

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Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>

<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>

<em />

I hope it helps!

5 0
3 years ago
Can you Please answer this question?
olga2289 [7]

Answer:

first option is not true

Explanation:

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3CO2 has 3 moles so 3 × 6.02 × 10²³ = 1.806 × 10²⁴ particles

4H2O has 4 moles so 4 × 6.02 × 10²³ = 2.408 × 10²⁴ particles

8 0
2 years ago
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