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3 years ago

How many half-lives have passed when there are 3 times as much daughter isotope as parent isotope?

Chemistry

1 answer:

Citrus2011 [14]3 years ago

3 0

Answer:

2 half lives.

Explanation:

Suppose there are 100g of parent isotope at the start.

After 1 half-life there will be 50g of parent and 50g of daughter isotope.

After another half life there is 25 g of parent and 75g of daughter isotope.

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Calculate the mass of MgCO3 precipitated by mixing 10.00 mL of a 0.200 M Na2CO3 solution with 5.00 mL of a 0.0450 M Mg(NO3)2 sol

aliya0001 [1]

Answer:

1.90 g

Explanation:

When we are making calculations based on chemical reactions, we need first to have a balanced chemical equation.

From there we can determine the mass of the product MgCO₃ in this question.

Na₂CO₃ + Mg(NO₃)₂ ⇒ 2 NaNO₃ + MgCO₃ ( double decomposition )

0.200 M 0.0450 M ?

10.0 5.00 mL

Now we know the volume and concentration of Mg(NO₃)₂,and Na₂CO₃ so we must compute their number of moles to determine the limiting reagent, if any.From there determine moles and mass of MgCO₃ produced.

First lets convert the volume of Mg(NO₃)₂and Na₂CO₃ to liters:

5.00 mL x ( 1 L/1000 mL ) = 5.00 x 10⁻³ L

10.00 mL x ( 1L/ 1000 mL ) = 1.000 x 10 ⁻² L

# mol Mg(NO₃)₂ = ( 0.0450 mol /L ) x 5.00 x 10⁻³ L

= 2.25 x 10⁻⁴ mol Mg(NO₃)₂

# mol Na₂CO₃ = ( 0.200 mol / L ) x 1 x 10⁻² L

= 2.000 x 10⁻³ mol Na₂CO₃

Calculation limiting reagent:

= 2.25 x 10⁻⁴ mol Mg(NO₃)₂ x ( 1 mol Na₂CO₃ / mol Mg(NO₃)₂ )

= 2.25 x 10⁻⁴ mol Na₂CO₃ required to react

Therefore, our limiting reagent is Mg(NO₃)₂ since we require 2.25 x 10⁻⁴ mol Na₂CO₃ to react completely with 2.25 x 10⁻⁴Mg(NO₃)₂, and we have excess of it.

# mol MgCO₃ produced

= 2.25 x 10⁻⁴ mol Mg(NO₃)₂ x ( 1 mol MgCO₃ / 1 mol Mg(NO₃)₂ )

= 2.25 x 10⁻⁴ mol MgCO₃

Now that we have the moles of MgCO₃, we can obtain its mass by multiplying its molar mas ( 84.31 g/mol ):

2.25 x 10⁻⁴ mol MgCO₃ x 84.31 g/mol = 1.90 g

7 0

3 years ago

In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.

olga nikolaevna [1]

Answer:

a. The second run will be faster.

d. The second run has twice the surface area.

Explanation:

The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.

$A=\pi ^{1/3} (6V)^{2/3}$

The area of the 10.0 cm³-sphere is:

$A=\pi ^{1/3} (6.10.0)^{2/3}=22.4cm^{2}$

The area of each 1.25 cm³-sphere is:

$A=\pi ^{1/3} (6. 1.25)^{2/3}=5.61cm^{2}$

The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²

The ratio of 8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00

Since the surface area is doubled, the second run will be faster.

6 0

3 years ago

Convert 3.01€22 molecules of O²(a) mole of O²) ,volume at S.T.P,no of Oxygen atom. (d) mass

Marat540 [252]

Answer:

a )0.05 moles of O₂

b) 1.1207 dm³

c) 0.3× 10²³ atoms of oxygen

d) 1.6 g

Explanation:

Given data:

Number of molecules of O₂ = 3.01 × 10²² molecules

Number of moles of O₂ = ?

Volume of oxygen at ATP = ?

Number of oxygen atoms = ?

Mass of oxygen = ?

Solution:

The given problem will solve by using Avogadro number.

"It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance". The number 6.022 × 10²³ is called Avogadro number.

Moles of oxygen:

1 mole = 6.022 × 10²³ molecules

3.01 × 10²² molecules × 1 mol / 6.022 × 10²³ molecules

0.05 moles of O₂

Number of atoms of oxygen:

1 mole = 6.022 × 10²³ atoms

0.05 mol × 6.022 × 10²³ atoms /1 mol

0.3× 10²³ atoms of oxygen

Volume of oxygen:

1 mole of oxygen at STP occupy 22.414 dm³

0.05 mol × 22.414 dm³ / 1mol

1.1207 dm³

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.05 mol × 32 g/mol

Mass = 1.6 g

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3 years ago

Element and symbol for bismuth

ahrayia [7]

Hello! Bismuth is a rainbowed colored metal which has the chemical symbol Bi and has an atomic number of 83 :D

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3 years ago

Is a neutron located in the inside or outside of the nucleus ?

vampirchik [111]

The neutron is inside the nucleus

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