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Alona [7]
4 years ago
14

There are several technological applications for the transuranium elements (Z > 92). An important one is in smoke detectors,

which can use the decay of a tiny amount of americium-241 to neptunium-237.What subatomic particle is emitted from that decay process? Write a balanced nuclear equation for this process.
Chemistry
1 answer:
timama [110]4 years ago
7 0

Answer:

An alpha particle is emitted.

95241Am → 42α + 23793Np

Explanation:

If we look at the difference between the atomic masses of the Americium isotope and neptunium isotope, we see a difference of 4 I e 241 - 237 = 4

Also,if we look at the difference between the atomic numbers of both nuclei, we get a difference of 2 I.e 95 - 93 = 2

Both differences show the exact properties of an alpha decay.

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4vir4ik [10]
Bonjeur monseur, la french are here! it seems you have a problem with the elements on earth. lead seems correct since it is indeed a harmful material
8 0
4 years ago
An iceberg has a volume of 9470 ft3.What is the mass in kilograms of the iceberg? The density of ice in the iceberg is 0.92 g/cm
Gemiola [76]
To determine the mass of the iceberg, we simply multiply the density with the volume. Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. Remember that the units should be homogeneous. We calculate as follows:

mass = density x volume
mass = 0.92 g /cm^3 ( 9470 ft^3) ( 30.48 cm / 1 ft )^3 ( 1 kg / 1000 g ) = 265.55 kg
6 0
4 years ago
Nuclear decay occurs according to first-order kinetics. What is the half-life of europium-152 if a sample decays from 10.0 g to
djyliett [7]

Answer:

13.5 years

Explanation:

Initial Concentration [Ao] = 10g

Final Concentration [A] = 0.768g

Time t= 50 years

Half life t1/2 = ?

These quantities are related by the following equations;

ln[A] = ln[Ao] - kt ......(i)

t1/2 = ln(2) / k ...........(ii)

where k = rate constant

Inserting the values in eqn (i) and solving for k, we have;

ln(0.768) = ln(10) - k (50)

-0.2640 = 2.3026 - 50k

50k = 2.3026 + 0.2640

k = 2.5666 / 50 = 0.051332

Insert the value of k in eqn (ii);

t1/2 = ln(2) / k

t1/2 = 0.693 / 0.051332 = 13.5 years

4 0
3 years ago
Serve(s) to increase cellular surface area.
omeli [17]
The answer is microvilli.
7 0
3 years ago
Read 2 more answers
Predict the pressure that would be observed if the container was at a temperature of 1,135 K.
denis-greek [22]

This problem is asking to predict the pressure in the container at a temperature of 1,135 K with no apparent background; however, in similar problems we can be given a graph having the pressure on the y-axis and the temperature on the x-axis and a trendline such as on the attached file, which leads to a pressure of 21.2 atm by using the given equation and considering the following:

<h3>Graph analysis.</h3>

In chemistry, experiments can be studied, modelled and quantified by using graphs in which we have both a dependent and independent variable; the former on the y-axis and the latter on the x-axis.

In addition, when data is recorded and graphed, one can use different computational tools to obtain a trendline and thus, attempt to find either the dependent or independent value depending on the requirement.

In this case, since the provided trendline by the graph and the program it was put in is y = 0.017x+1.940, we understand y stands for pressure and x for temperature so that we can extrapolate this equation even beyond the plotted points, which is this case.

In such a way, we can plug in the given temperature to obtain the required pressure as shown below:

y = 0.017 ( 1,135 ) + 1.940

y = 21.2

Answer that is in atm according to the units on the y-axis:

Learn more about trendlines: brainly.com/question/13298479

7 0
3 years ago
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