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3 years ago

If a substance has a half life of 58 years and starts with 500 g radioactive, how much remains radioactive after 30 years?

Chemistry

1 answer:

Vilka [71]3 years ago

5 0

Answer:

A = 349 g.

Explanation:

Hello there!

In this case, since the radioactive decay kinetic model is based on the first-order kinetics whose integrated rate law is:

$A=Ao*exp(-kt)$

We can firstly calculate the rate constant given the half-life as shown below:

$k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{58year}=0.012year^{-1}$

Therefore, we can next plug in the rate constant, elapsed time and initial mass of the radioactive to obtain:

$A=500g*exp(-0.012year^{-1} *30year)\\\\A=349g$

Regards!

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2 years ago

An oxygen atom has a mass of 2.66 x 10^-23 g and a glass of water has a mass of 0.050kg. Use this information to answer the ques

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<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>

Explanation:

We know, each mole contains $N_A=$ $6.023 \times 10^{23}$ atoms.

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Therefore, mass of one mole of oxygen, $M=m\times N_A$ .

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Mass of water in glass=0.050 kg = 50 gm.

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2 years ago

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3 years ago

What are some of the positive and<br> negative effects of the use of plastics?

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3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

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2 years ago