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Elan Coil [88]
2 years ago
5

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea

ding of 400. counts has diminished to 100. counts after 90.3 minutes , what is the half-life of this substance? Express your answer with the appropriate units. View Available Hint(s) t1/2 t 1 / 2 t_1/2 = nothing
Chemistry
1 answer:
taurus [48]2 years ago
8 0

Answer:

t₁/₂ = 45.1 min

Explanation:

The  radioactive decay equation is given by

N/N₀ = e^-kt  where N = counts after time t

                                 N₀ = counts initially

                                 k = decay constant

                                 t=  time elapsed

The question is what´s the half-life of this substance, and we can solve it once we have k from the expression above since

                                 t₁/₂ = 0.693/k

which is derived from  that equation, but for the case    N/N₀ is 0.5

Lets calculate k and  t₁/₂ :

N/N₀ = e^-kt      (taking ln in the two sides of the equation)

ln (N/N₀) =  ln e^-kt   = -kt   ⇒ k = -ln(N/N₀)/t

k = -ln(100/400)/90.3 min = 0.01535 min⁻¹

 t₁/₂ = 0.693/k  = 0.693/0.01535 min⁻¹  = 45.1 min

We can check this answer since the time in the question is the double of this  half-life and the data shows the material has decayed by a fourth: two half-lives.

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Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(
kkurt [141]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2Sr(s)+O_2(g)\rightarrow 2SrO(s)    \Delta H_1=-1184kJ

(2) SrO(s)+CO_2(g)\rightarrow SrCO_3(s)     \Delta H_2=-234kJ      ( × 2)

(3) CO_2(g)\rightarrow C(s)+O_2(g)     \Delta H_3=394kJ    ( × 2)

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ

Hence, the \Delta H^o_{rxn} for the reaction is 72 kJ.

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Answer the question PLS 100 points!!!
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2 years ago
Ethanoic acid, CH3COOH, ionizes to form an ethanoate ion (CH3COO-) in aqueous solution. what else does it form? is it a monoprot
barxatty [35]

Ethanoic acid ionizes in aqueous solutions to form two ions which are CH_3COO^- and H^+

<h3>Ionization of ethanoic acid</h3>

Ethanoic acid goes by the chemical formula CH_3COOH.

In aqueous solutions, it ionizes as a monoprotic acid according to the following equation:

CH_3COOH ---- > CH_3COO^- + H^+

A monoprotic acid is an acid that is able to donate only a proton. Hence, ethanoic acid is said to be monoprotic because it ionizes in aqueous solutions to produce a single H^+

More on ethanoic acid can be found here: brainly.com/question/9991017

#SPJ1

8 0
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