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SSSSS [86.1K]
3 years ago
6

The barium isotope 133Ba has a half-life of 10.5 years. A sample begins with 1.1×1010 133Ba atoms. How many are left after (a) 6

years, (b) 10 years, and (c) 200 years?
Chemistry
1 answer:
Svetach [21]3 years ago
6 0

Answer:

(a) 7.4 x 10⁹ atoms.

(b)

(c)

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
  • The half-life of 133-Ba = 10.5 years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(10.5 years) = 0.066 year⁻¹.

  • Also, we have the integral law of first order reaction:

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 0.066 year⁻¹).

t is the time of the reaction.

[A₀] is the initial concentration of (133-Ba) ([A₀] = 1.1 x 10¹⁰ atoms).

[A] is the remaining concentration of (133-Ba) ([A] = ??? g).

<u><em>(a) 6 years:</em></u>

t = 6.0 years.

∵ kt = ln([A₀]/[A])

∴ (0.066 year⁻¹)(6.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])

∴ 0.396 = ln((1.1 x 10¹⁰ atoms)/[A]).

<em>Taking exponential for both sides:</em>

∴ 1.486 = ((1.1 x 10¹⁰ atoms)/[A]).

∴ [A] = (1.1 x 10¹⁰ atoms)/(1.486) = 7.4 x 10⁹ atoms.

<u><em>(b) 10 years</em></u>

t = 10.0 years.:

∵ kt = ln([A₀]/[A])

∴ (0.066 year⁻¹)(10.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])

∴ 0.66 = ln((1.1 x 10¹⁰ atoms)/[A]).

<em>Taking exponential for both sides:</em>

∴ 1.935 = ((1.1 x 10¹⁰ atoms)/[A]).

∴ [A] = (1.1 x 10¹⁰ atoms)/(1.935) = 5.685 x 10⁹ atoms.

<u><em>(c) 200 years:</em></u>

t = 200.0 years.

∵ kt = ln([A₀]/[A])

∴ (0.066 year⁻¹)(200.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])

∴ 13.2 = ln((1.1 x 10¹⁰ atoms)/[A]).

<em>Taking exponential for both sides:</em>

∴ 5.4 x 10⁵ = ((1.1 x 10¹⁰ atoms)/[A]).

∴ [A] = (1.1 x 10¹⁰ atoms)/(5.4 x 10⁵) = 2.035 x 10⁴ atoms.

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guajiro [1.7K]
It should be 8 O atoms. 3O atoms in Na2S2O3 and 5O atom in 5H2O.  The reason there are 5 O atoms are because the 5 in front of H2O means you multiply each atom in the compound by that number (like the distributive property). The H2 molecule becomes 10 Hydrogen atoms (5*2) and the Oxygen becomes 5 Oxygen atoms (5*1). Then you add the 5O atoms to the 3O atoms which equals 8
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Molar mass (NH4)2SO4
Hoochie [10]

Answer:

<u>132.15</u>

Explanation:

Molar mass N = 14.00

Molar mass H = 1.01

Molar mass H4 = 1.01 x 4 = 4.04

Molar mass NH4 = 14.00 + 4.04 = 18.04

Molar mass (NH4)2 = 18.04 x 2 = 36.08

Molar mass S = 32.07

Molar mass O = 16.00

Molar mass O4 = 16.00 x 4 = 64.00

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Molar mass (NH4)2SO4 = 36.08 + 96.07 = <u>132.14</u>

6 0
3 years ago
According to the chart of specific heats, which material requires the most heat to raise its temperature from 10°C to 30°C?
myrzilka [38]

The material which requires the most heat to raise its temperature from 10°C to 30°C is oil.

<h3>What is the formula to calculate absorbed heat?</h3>

The formula which we used to calculate the amount of involved heat in relation with specific heat is:

Q = mcΔT, where

  • Q = absorbed heat
  • m = mass
  • c = specific heat
  • ΔT = change in temperature

Among the given materials, specific heat of oil is highest than other materials so will require maximum absorbed heat.

Hence, oil requires the most heat.

To know more about specific heat, visit the below link:

brainly.com/question/6198647

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Increase the pressure on the system
Molodets [167]

Answer:

What do you mean????????

6 0
3 years ago
46.6 grams of mercury II sulfate (HgSO4) reacts with an excess of sodium Chloride (NaCl). How many grams of mercury II chloride
slega [8]

Answer:

m_{HgCl_2}=42.7gHgCl_2

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

HgSO_4+2NaCl\rightarrow HgCl_2+Na_2SO_4

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

m_{HgCl_2}=46.6gHgSO_4*\frac{1molHgSO_4}{296.65 gHgSO_4}*\frac{1molHgCl_2}{1molHgSO_4} *\frac{271.52gHgCl_2}{1molHgCl_2} \\\\m_{HgCl_2}=42.7gHgCl_2

Best regards.

3 0
3 years ago
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