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3 years ago

Could you determine the half life of an isotope by looking at 1 atom? Explain why or why not.

Chemistry

2 answers:

Katyanochek1 [597]3 years ago

7 0

Answer:

single, antom-radioactive isotope U-238.

Explanation:

Bezzdna [24]3 years ago

3 0

Answer:

Hope this helps u!!

Explanation:

A useful application of half-lives is radioactive dating. This has to do with figuring out the age of ancient things. If you could watch a single atom of a radioactive isotope, U-238, for example, you wouldn't be able to predict when that particular atom might decay. :) :)

Mark as Brainliest plz!

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What would be the formula for an alkane which contains 30 hydrogen atoms? C__H30

maxonik [38]

C14H30 because of the formula for alkanes:
CnC2n+2

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3 years ago

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How many grams of sulfuric acid are needed to produce 57.0 grams of water? Show all steps of your calculation as well as the fin

antiseptic1488 [7]

Answer:

155.16 g.

Explanation:

Firstly, It is considered as a stichiometry problem.

From the balanced equation: 2NaOH + H₂SO₄ → 2Na₂SO₄ + 2H₂O

It is clear that the stichiometry shows that 2.0 moles of NaOH reacts with 1.0 mole of H₂SO₄ to give 2.0 moles of Na₂SO₄ and 2.0 moles of H₂O.

We must convert the grams of water (57.0 g) to moles (n = mass/molar mass).

n = (57.0 g) / (18.0 g/mole) = 3.1666 moles.

Now, we can get the number of moles of H₂SO₄ that is needed to produce 3.1666 moles of water.

Using cross multiplication:

1.0 mole of H₂SO₄ → 2.0 moles of H₂O, from the stichiometry of the balanced equation.

??? moles of H₂SO₄ → 3.1666 moles of H₂O.

The number of moles of H₂SO₄ that will produce 3.1666 moles of H₂O (57.0 g) is (1.0 x 3.1666 / 2.0) = 1.5833 moles.

Finally, we should convert the number of moles of H₂SO₄ into grams (n = mass/molar mass).

Molar mass of H₂SO₄ = 98.0 g/mole.

mass = n x molar mass = (1.5833 x 98.0) = 155.16 g.

6 0

3 years ago

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A 10 gram sample of H20 is sealed in a 1350 ml flask at 27°C. Given the fact that water has a vapor pressure of 26.7 mmHg at thi

Aleksandr-060686 [28]

Answer:

9.9652g of water

Explanation:

The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:

n = PV / RT

Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)

Replacing:

n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K

n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:

1.93x10⁻³ moles × (18.01g / 1mol) = 0.0348g of water

As the initial mass of water was 10g, the mass of water that remains in liquid phase is:

10g - 0.0348g = 9.9652g of water

I hope it helps!

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3 years ago

When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0

neonofarm [45]

Answer:

The mass of PbSO4 formed 15.163 gram

Explanation:

mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625

mole of Na₂SO₄ = 2 x 0.025 = 0.05

Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃

( Mole/Stoichiometry ) $\frac{0.0625}{1}$ $\frac{0.05}{1}$

= 0.0625 = 0.05

From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.

Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄

= 0.05 x 303.26 g

= 15.163 g

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3 years ago

Chiara knows that weight is affected by gravitational pull. She is putting together a poster to display in her classroom.

Tanya [424]

Answer: A

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8 0

3 years ago

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