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muminat
2 years ago
11

A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d

ecay rate for 1.000 mole of this nuclide. decays s-1
Chemistry
1 answer:
Karo-lina-s [1.5K]2 years ago
8 0

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

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Calculate the density of an object with a mass of 220 and a volume of 145mL.
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Answer:

1.52 g/mL

Explanation:

The formula for finding the density of an object is d = m/v, or density = mass divided by volume.

Therefore, we can input the two values and solve for d.

d = 220/145

d = 44/29

d ≈ 1.52 g/mL

8 0
2 years ago
How many moles are in 2.8 x 1023 BS3 molecules?
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Answer:

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What happens to the motion of molecules in a substance is cooled?
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2 years ago
A ballon is inflated with 2.42L of helium at a temperature of 27.0°C. When put in the freezer, the volume changes to 2.37L and -
natita [175]

Answer:

838 torr  

Step-by-step explanation:

To solve this problem, we can use the <em>Combined Gas Laws</em>:

p₁V₁/T₁ = p₂V₂/T₂            Multiply each side by T₁

   p₁V₁ = p₂V₂ × T₁/T₂     Divide each side by V₁

      p₁ = p₂ × V₂/V₁ × T₁/T₂

<em>Data: </em>

p₁ = ?;             V₁ = 2.42 L; T₁ =  27.0 °C

p₂ = 754 torr; V₂ = 2.37 L; T₂ =  -8.8 °C

Calculations:

(a) Convert <em>temperatures to kelvins </em>

T₁ = (27.0 + 273.15) K = 300.15 K

T₂ = (-8.8 + 273.15) K = 264.35 K

(b) Calculate the<em> pressure </em>

p₁ = 754 torr × (2.37 L/2.42) × (300.15/264.35)  

p₁ = 754 torr × 0.979 × 1.135

p₁ = 838 torr

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3 years ago
If a chemical reaction produces 20.0 grams of product, but by stoichiometry it is supposed to have 25.0 grams of product; what i
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