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Romashka [77]
2 years ago
11

Sodium-25 was to be used in an experiment, but it took 3.0 minutes to get the sodium from the reactor to the laboratory. If 8.0

mg of sodium-25 was removed from the reactor, how many mg of sodium-25 were placed in the reaction vessel 3.0 minutes later if the half life of sodium-25 is 60 seconds.
Chemistry
1 answer:
Ksenya-84 [330]2 years ago
5 0

Sodium-25 after 3 minutes : 1.0625 mg

<h3>Further explanation</h3>

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

<h3 />

No=8 mg

t1/2=60 s

T=3 min=180 s

\tt Nt=No\dfrac{1}{2}^{T/t1/2}\\\\Nt=8.5\dfrac{1}{2}^{180/60}\\\\Nt=8.5(\dfrac{1}{2})^3\\\\Nt=1.0625~mg

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Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g o
konstantin123 [22]

Answer:

- From octane: m_{CO_2}=1.54x10^5gCO_2

- From ethanol: m_{CO_2}=9.57x10^4gCO_2

Explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O

Thus, the produced mass of carbon dioxide is:

m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2

Now, for ethanol:

C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O

m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2

Best regards.

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