A 50 gram sample of radioisotope undergoes 2 half lives. How many grams would remain at the end?

When optically active (S)-2-methylcyclopentanone is treated with aqueous base, the compound loses its optical activity. Explain

Sunny_sXe [5.5K]

Answer:

See explanation below

Explanation:

In this case, we need to see which is the structure of this compound. Now, racemization occurs basically because we are in an aqueous basic medium, and the ketone can reacts again with water in the medium to form the starting reagent.

First, the base will take out the Alpha hydrogen from the ketone, then, the negative charge goes down and opens up the carbonile group, forming a double bond in there. Later, with the water of the medium, it reacts and substract a proton, and then, with keto enolic equilibrium, forms again the ketone, but this ketone is different from the start, it will be the R isomer which is not optically active.

See picture below for mechanism

3 0

2 years ago

When does carbon dioxide absorb the most heat energy?

Lyrx [107]

The correct answer of the given question above would be option C. Carbon dioxide absorbs the most heat energy during sublimation. It is the<span> transition of a substance directly from the solid to the gas phase without passing through the intermediate liquid phase. Hope this answers your question.</span>

5 0

2 years ago

Read 2 more answers

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0

3 years ago

A 10 gram sample of iron reacts with oxygen to form 18.2 grams of ferric oxide. How many grams of oxygen reacted?

RoseWind [281]

Answer:

$\boxed {\boxed {\sf 8.2 \ grams}}$

Explanation:

According to the Law of Conservation of Mass, the mass of the products must equal the mass of the reactants.

mass products = mass reactants

In this problem, the reaction is:

$iron + oxygen = ferric \ oxygen$

The reactants are iron and oxygen. We know the mass of the iron sample is 10 grams.

The product is ferric oxide. The mass of the ferric oxide sample is 18.2 grams.

$10 \ g + oxygen=18.2 \ g$

We want to find how many grams of oxygen reacted. We have to get the oxygen by itself. 10 is being added to oxygen. The inverse of addition is subtraction. Subtract 10 from both sides of the equation.

$10 \ g - 10 \ g+ oxygen = 18.2 \ g - 10 \ g$