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Alex_Xolod [135]
3 years ago
11

The isotope 208Tl undergoes β decay with a half-life of 3.1 min.

Chemistry
1 answer:
melamori03 [73]3 years ago
8 0

Answer:

a. Pb 208

b. About 21.7 minutes

c. only a trace amount

Explanation:

It under goes beta decay.

There should be virtually nothing after an hour

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Now we find out how many each unit is:
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Now we multiply it by 3 to find the number of moles of O2
3(6) = 18 mol O2. 
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C(graphite) + O2(g) → CO2(g) + 94.05 kcal C(diamond) + O2(g) → CO2(g) + 94.50 kcal What could you infer from the information giv
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Looking at both reactions, we can see that the combustion of carbon, graphite or diamond, leads to the formation of carbon dioxide and energy. Since energy is a product of the reaction, we known that this is an exothermic reaction. The value of the change in enthalpy, ΔH, will be negative.

A negative value of ΔH in each exothermic reaction suggests that the carbon dioxide product is at a lower energy than the reactants. And since more energy is released in the combustion of diamond compared to graphite, we know that diamond has a higher internal energy than graphite.
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3 years ago
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An electromagetic wave can only move through _____. a Air b Water c Empty space d Land
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The answer is a) air
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What are the primary products of burning a fossil fuel?
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Hydrocarbons are a broad class of pollutants made up of hundreds of specific compounds containing carbon and hydrogen. The simplest hydrocarbon, methane, does not readily react with nitrogen oxides to form smog, but most other hydrocarbons do. Hydrocarbons are emitted from human-made sources such as auto and truck exhaust, evaporation of gasoline and solvents, and petroleum refining.

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5 0
3 years ago
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N2+3H2 → 2NH3
s2008m [1.1K]

Explanation:

N2 (g) + H2 (g) gives out NH3 (g)

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.(If you balance the hydrogen reactant with a whole number first, I can guarantee you that you will have to give NH3 a new stoichiometric coefficient.)

N2 (g) + 3H2 (g) gives out 2NH3 (g)

The stoichiometric coefficients tell you that if we can somehow treat every component in the reaction as the same (like on a per-mol basis, hinthint), then one "[molar] equivalent" of nitrogen yields two [molar] equivalents of ammonia.

Luckily, one mol of anything is equal in quantity to one mol of anything else because the comparison is made in the units of mols.

So what do we do? Convert to

mols (remember the hint?).

28g N2 × 1 mol N2/ 2 × 14.007) g N2

= 0.9995 mol N2

At this point you don't even need to calculate the number of mols of H2 . Why? Because H2 is about 2 g/mol, which means we have over 10 mols of H2. We have 1 mol N2, and we need three times as many mols of H2 as we have

N2.

After doing the actual calculation you should realize that we have about 4 times as much H2 as we need. Therefore the limiting reagent is clearly N2.

Thus, we should yield 2×0.9995=1.9990 mols of NH3 (refer back to the reaction). So this is the second and last calculation we need to do:

1.9990 mol NH3 × 17.0307 g NH3/ 1 mol NH3

= 34.0444 g NH3

Hope it helpz~

4 0
2 years ago
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