Calculate the area of a floor that is 1.5 × 103 m long and 2.2 × 102 m wide.

How much of a radioactive parent isotope will remain after three half-lives have passed?

Sedaia [141]

After three half lives have passed, there would be only 12.5 percent of the original amount of a radioactive parent isotope that will remain. Half life is the time needed for a certain amount of a substance to be half its initial amount. It is a common term used in nuclear chemistry describing how fast radioactive substances undergo decay. One half life would correspond to only 50% would be left. Two half lives would be 25% only of the original value. Three half lives would be 12.5%. Four half lives would be 6.25% of the initial value. So on and so forth.

4 0

3 years ago

You have placed a can of lemonade in a cooler of is ice.what is the best way to describe what is happening to the ice

Firdavs [7]

<span>Heat from the can is making the water molecule move faster, which melts the ice.</span>

6 0

3 years ago

Read 2 more answers

A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas

sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

CaCO3 + H2SO4 → CaSO4 + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4 × $\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}$ $\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}$

= 21284.56606

mass of CaCO3 = 2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

brainly.com/question/12498769

#SPJ4

4 0

1 year ago

What is the name of the molecular compound CBra?

Fudgin [204]

Answer:

i think it is a Cobrotoxin

6 0

2 years ago

What were the reslt of aurangzeb's religion policy ? what measures did he takes

soldier1979 [14.2K]

Answer: The religious fanaticism of Aurangzeb overshadowed his virtues. His reversal of Akbar's policy of religious toleration resulted in weakening the entire structure of the Mughal empire. It led to several conflicts and wars in different parts of the country.

5 0

3 years ago