CHEMISTRY HIGH SCHOOL

A radioactive isotope has a half-life of 10 minutes.At the start of an experiment, the activity of a sample of this isotope was 800 counts
per second after allowing for background radiation.
Calculate how long it would be before the activity fell from 800 counts per second to
200 counts per second.

Answers

Answer 1
Answer:

Answer: The time for activity to fell from 800 counts per second to 200 counts per second is 20 minutes

Explanation:

Expression for rate law for first order kinetics is given by:

t=(2.303)/(k)\log(a)/(a-x)

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for finding the rate constant:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{(1)/(2)}=(0.693)/(k)

k=(0.693)/(10min)=0.0693min^(-1)

b) for activity to fell from 800 counts per second to 200 counts per second.

t=(2.303)/(0.0693)\log(800)/(200)

t=20minutes

The time for activity to fell from 800 counts per second to 200 counts per second is 20 minutes
Answer 2
Answer:

Final answer:

The half-life of the given radioactive isotope is 10 minutes. The activity of the isotope falls to half its initial value every 10 minutes. Hence, it would take 20 minutes for the activity to fall from 800 counts per second to 200 counts per second.

Explanation:

The process you're describing relates to the concept of radioactive decay, which is a fundamental principle in nuclear physics. In this case, we're dealing with what's known as a half-life. The half-life of a radioactive isotope is the time it takes for half of the isotope in a sample to decay.

In this scenario, the radioactive isotope has a half-life of 10 minutes and it starts with a sample activity of 800 counts per second. According to the nature of half-life, it will reduce to half of its initial counts, 400 counts per seconds in 10 minutes. Importantly, another 10 minutes would be required for the number of counts to halve again to 200 counts per second.

Therefore, it would take 20 minutes for the activity to fall from 800 counts per second to 200 counts per second.

Learn more about Radioactive Decay here:

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Which element has the same number of energy levels as krypton (Kr) and the same number of valence electrons as nitrogen (N) ? iron (Fe)
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Answers

Answer:

The correct answer is the option: arsenic (As).

Explanation:

Hello!

Let's solve this!

Nitrogen is in group V period 2 and has 5 valence electrons.

The krypton is in period 4.

If we cross these two elements, the element that is in the same group as nitrogen (V) and in the same period as Kripton (4) is Arsenic. If you have any questions, we can check it on the periodic table.

We conclude that the correct answer is the option: arsenic (As).

Answer:

The answer to your question is: Arsenic (As)

Explanation:

Krypton is a noble gas located in group VIII A and 4rd period. Nitrogen is in the 5fh group.

So, we must look for an element located in the 4rd period and group VA.

Iron (Fe)               Iron is in the 4rd period and group VI B

Selenium (Se)     Selenium is in the 4rd period and group VI A

Arsenic (As)       Arsenic is in the 4rd period and group V A and group VA

Calcium (Ca)     Calcium is located in group 4rd and group IIA

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The process J is called deposition. Using water is an example; write an equation that represents the process of deposition.​

Answers

In chemistry and phase equilibria, phase transition is a common phenomenon. There are many terms related to the phase transition. The process deposition is a physical process. This process is the opposite of sublimation.

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An example is:

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Which of the statement below states the negative impact did chemistry have on society through the production and use of chlorofluorocarbons (CFCs)? A. Hazardous coolants in refrigeration units were replaced. B. Better refrigeration reduced loss of food through spoilage. C. Lower levels of ozone allowed more ultraviolet radiation to reach Earth. D. Fire safety improved through the use of fire extinguishers.

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Answer:

C. Lower levels of ozone allowed more ultraviolet radiation to reach Earth.

Explanation:

got it right on edge 23

How many grams of magnesium oxide can be produced when 97.2 g Mg react with 88.5 g O2?

Answers

First we establish the chemical reaction equation. 
2Mg + O2 = 2MgO. 
Then we need to know which of the reactant is limiting. We do that by multiplying the number of moles of reactant to the stoichiomeric coefficient. For Magnesium: 97.2 g* (mol)/(24g) *2 = 8
For O2: 88.5g*(mol)/(32g) *1 = 2.7
Since O2 has the smaller moles, this is the Limiting reactant. Then we solve basing on the number of O2 used. 

88.5g* (mol)/(32g) * (2)/(1) * (56g)/(mol) = 309.75 g

Answer : The mass of magnesium oxide produced will be, 161.2 g

Solution : Given,

Mass of Mg = 97.2 g

Mass of O_2 = 88.5 g

Molar mass of Mg = 24.3 g/mole

Molar mass of O_2 = 32 g/mole

Molar mass of MgO = 40.3 g/mole

First we have to calculate the moles of Mg and O_2.

\text{Moles of Mg}=\frac{\text{Mass of Mg}}{\text{Molar mass of Mg}}=(97.2g)/(24.3g/mole)=4\text{ moles}

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=(88.5g)/(32g/mole)=2.7\text{ moles}

The balanced reaction is,

2Mg+O_2\rightarrow 2MgO

As, 2 moles of Mg react with 1 moles of O_2

So, 4 moles of Mg react with (4)/(2)=2 moles of O_2

From this we conclude that the O_2 is in excess amount and Mg is in limited amount.

Now from the reaction we conclude that

As, 2 moles of Mg react to give 2 moles of MgO

So, 4 moles of Mg react to give 4 moles of MgO

Now we have to calculate the mass of MgO.

\text{Mass of MgO}=\text{Moles of MgO}* \text{Molar mass of MgO}

\text{Mass of MgO}=(4moles)* (40.3g/mole)=161.2g

Therefore, the mass of magnesium oxide produced will be, 161.2 g

If the mass of a material is 111 grams and the volume of the material is 25 cm3, what would the density of the material be? g/cm3

Answers

86 it lit like bic bihh!!!!!!!!!!!!!!!!!!!!!
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