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ale4655 [162]
2 years ago
8

Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio

of the mass of 14 6c to the total mass of carbon in the atmosphere is r, and the decay constant of 14 6c is λ. assume that, at any time, 14 6c is a negligible fraction of the total mass of carbon and that the measured activity of the sample is purely due to 14 6c. also assume that the ratio of mass of 14 6c to total carbon mass in the atmosphere (the source of the carbon in the sample) is the same at present and on the day when the number of 14 6c atoms in the sample was set. express your answer in terms of the mass ma of a 14 6c atom, mc, a, r, and λ.
Chemistry
1 answer:
Paha777 [63]2 years ago
6 0
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
 The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) = 
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).

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How did Mendeleev feel about Russian science education when he came back from Germany
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In a laboratory experiment, the freezing point of an aqueous solution of glucose is found to be -0.325°C, What is the molal conc
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0.17 M is the is the molal concentration of this solution

Explanation:

Data given:

freezing point of glucose solution = -0.325 degree celsius

molal concentration of the solution =?

solution is of glucose=?

atomic mass of glucose = 180.01 grams/mole

freezing point of glucose = 146 degrees

freezing point of water = 0 degrees

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2 years ago
Write a balanced equation for the transmutation that occurs when a scandium-48 nucleus undergoes beta decay.
Tju [1.3M]

Answer:

A scandium-48 nucleus undergoes beta-minus decay to produce a titanium-48 nucleus.

\rm ^{48}_{21}Sc \to ^{48}_{22}Ti + ^{\phantom{1}\,0}_{-1}e^{-} + \bar{\mathnormal{v}}_e.

Explanation:

There are two types of beta decay modes: beta-minus and beta-plus.

In both decay modes, the mass number of the nucleus stays the same.

However, in a beta-minus decay, the atomic number of the nucleus increases by one. In a beta-plus decay, the atomic number decreases by one.

Each beta-minus decay releases one electron and one electron antineutrino. Each beta-plus decay releases one positron and one electron neutrino.

Look up the atomic number and relative atomic mass for the element scandium.

  • The atomic number of \rm Sc is 21.
  • The relative atomic mass of \rm Sc is approximately 45.0.

This question did not specify whether the decay here is beta-plus or a beta-minus. However, the relative atomic mass of this element can give a rough estimate of the mode of decay.

Each element (e.g, \rm Sc) can have multiple isotopes. These isotopes differ in mass. The relative atomic mass of an element is an average  across all isotopes of this element. This mass is weighted based on the relative abundance of the isotopes. Its value should be closest to the most stable (and hence the most abundant) isotope.

The mass number of scandium-48 is significantly larger than the relative atomic mass of this element. In other words, this isotope contains more neutrons than isotopes that are more stable. There's a tendency for that neutron to convert to a proton- by beta-minus decay, for example.

The atomic number of the nucleus will increase by 1. 21 + 1 = 22. That corresponds to titanium. The mass number stays the same at 48. Hence the daughter nucleus would be titanium-48. Note that two other particles: one electron and one electron \rm e^{-} and one antineutrino \bar{v}_{\text{e}} (note the bar.) The neutrino helps balance the lepton number of this reaction.

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The table shows columns that Brenda uses for her notes on the properties of elements. Her notes state that some elements can rea
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Answer:

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An example of such properties that metals and metalloids have in common is that they have relative high melting points. Metalloids are all solid at room temperature, such as most metals.

Other property that both metals and metalloids share is that they can react with oxygen to form oxides that are amphoteric.

Amphoteric compounds are substances that can behave as a base or as an acid, depending on the other compound with which they react.

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The oxides of metalloids are usually amphoteric.

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