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klasskru
11 days ago
5

Two containers hold the same radioactive isotope. Container A contains 1000 atoms, and container B contains 500 atoms. Which of

the following statements about containers A and B is true? Two containers hold the same radioactive isotope. Container A contains 1000 atoms, and container B contains 500 atoms. Which of the following statements about containers A and B is true? The rate of decay of atoms (half-life) in container A is greater (or longer) than the rate of decay of atoms (half-life) in container B. The rate of decay of atoms (half-life) in container B is greater (or longer) than the rate of decay of atoms (half-life) in container A. The rate of decay of atoms (half-life) in container B is the same as the rate of decay of atoms (half-life) in container A.
Chemistry
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A 0.100 m solution of which one of the following solutes will have the highest vapor pressure? A 0.100 m solution of which one o
eduard [2782]
Vapor pressure refers to the force exerted by vapor or gas molecules above the surface of a liquid. It is inversely related to the concentration of solute particles; an increase in solute concentration results in a decrease in vapor pressure, and vice versa. For (a), it dissociates into two particles. In (b), the total count of particles from dissociation becomes 1 + 2, totaling three. For (c), dissociation yields 1 + 3 for a total of four particles. (d) Since sucrose is a covalent compound, it does not break apart into ions, so it remains as one particle. For (e), dissociation results in 1 + 1, equating to two particles.
5 0
2 months ago
Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t
lorasvet [2795]

Respuesta:

Un avión fabricado con aluminio puede transportar una mayor cantidad de pasajeros comparado con uno de acero.

Explicación:

La masa total que el avión es capaz de levantar es:

m_{tot}=m_{fuselage}+m_{passangers}

Para el aluminio:

m_{tot}=m_{fus-Al}+m_{pas-Al}

m_{fus-Al}=\delta _{Al}*V_{fuselage}

y

V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]

donde:

  • L es longitud
  • D es diámetro
  • e es grosor

m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}

Para el acero (mismo procedimiento):

m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel

Sabiendo que la masa total que el avión puede levantar es constante y que el aluminio tiene una densidad menor que la del acero, podemos afirmar que el avión de aluminio puede levantar un mayor número de pasajeros.

También es posible estimar un peso promedio de los pasajeros para calcular cuántos podría soportar.

5 0
4 months ago
A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a
KiRa [2933]

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g ×\frac{1mol}{159,609g} = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g ×\frac{1mol}{18,02g} = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

5 0
3 months ago
In living organisms, C-14 atoms disintegrate at a rate of 15.3 atoms per minute per gram of carbon. A charcoal sample from an ar
lions [2927]

Answer:

The result is "4,241.17 years"

Explanation:

The disintegration rate for C-14 atoms is indicated in  15.3 \frac{atoms}{min-g}

The dissolution rate of the sample is given by 9.16 \frac{atoms} {min-gram}

The C-14 proportion within the sample can be determined as per = \frac {9.16}{15.3} \\\\ = 0.5987

With a half-life of 5730 years.

Now, we need to compute the number of half-lives (n) that are applicable:

(\frac{1}{2})^n= A\\\\A= fraction of C-14, which is remaining \\\\(\frac{1}{2})^n= 0.5987 \\\\ n \log 2 = - \log 0.5987\\\\

\therefore \\\\ \Rightarrow n= \frac{0.227}{0.3010} \\\\ = 0.740\\

Thus, the age of the sample is represented as = n \times\ half-life

                                                 = 0.740 \times 5730 \ years \\\\=4241.17 \ years\\\\

6 0
3 months ago
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