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zhenek [66]
3 years ago
13

A 5-kg quantity of radioactive isotope decays to 2 kg after 10 years. Find the decay constant of the isotope.k = _____

Chemistry
1 answer:
Dvinal [7]3 years ago
7 0

Answer:

k = -0.09165 years^(-1)

Explanation:

The exponential decay model of a radioactive isotope is generally given as;

A(t) = A_o(e^(kt))

Where;

A_o is quantity of isotope before decay, k is decay constant and A(t) is quantity after t years

We are given;

A_o = 5 kg

A(10) = 2kg

t = 10 years

Thus;

A(10) = 2 = 5(e^(10k))

Thus;

2 = 5(e^(10k))

2/5 = (e^(10k))

0.4 = (e^(10k))

In 0.4 = 10k

-0.9164 = 10k

k = -0.9164/10

k = -0.09165 years^(-1)

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58×5/9

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10 (212-32)5/9

180×5/9

20×5

100°C

8 0
3 years ago
What is the name of the polyatomic in MgSO4?
blsea [12.9K]

Answer:

Magnesium Sulfate

Explanation:

4 0
3 years ago
Read 2 more answers
The heat of vaporization for benzaldehyde is 48.8 kj/mol, and its normal boiling point is 451.0 k. use this information to deter
user100 [1]

Answer:

The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

Explanation:

  • To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
  • The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
  • The second case: P₂ = <em>??? needed to be calculated</em> and T₂ = 61.5 °C = 334.5 K.
  • ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
  • Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)
  • ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)
  • ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.
  • (760 torr /P₂) = 0.01075
  • Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.

So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

7 0
3 years ago
How many L of a 7.5 H2SO4 stock solution would you need to prepare a dilute solution 100 L of 0.25M H2SO4
blsea [12.9K]

Answer:

3.33 L

Explanation:

We can solve this problem by using the equation:

  • C₁V₁=C₂V₂

Where the subscript 1 refers to one solution and subscript 2 to the another solution, meaning that in this case:

  • C₁ = 0.25 M
  • V₁ = 100 L
  • C₂ = 7.5 M
  • V₂ = ?

We input the data:

  • 0.25 M * 100 L = 7.5 M * V₂
  • V₂ = 3.33 L

Thus the answer is 3.33 liters.

3 0
2 years ago
How many mL of a 4% mass/volume Mg(NO3)2 solution would contain 1.2 grams of magnesium nitrate?
Phoenix [80]
4% mass / volume :

4 g ---------> 100 mL
1.2 g ------- ? mL

V = 1.2 * 100 / 4

V = 120 / 4

V = 30 mL

hope this helps!

7 0
3 years ago
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