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RoseWind [281]
3 years ago
15

When a dilute solution of 20minutes of H2O2 solution total volume of oxygen was collected was 120cm³ in just 15minutes. Calculat

e the rate of formation of oxygen.
Chemistry
1 answer:
Vanyuwa [196]3 years ago
7 0

Answer:

\large \boxed{\text{8.0 cm}^{3}\cdot \text{min}^{-1}}

Explanation:

The rate of formation of O₂ is the change in volume divided by the change in time.

For the reaction  

\rm 2H_{2}O_{2}\longrightarrow \, 2H_{2}O+ O_{2} \\\text{rate} = \dfrac{\Delta V}{\Delta t } = \dfrac{\text{120 cm}^{3}}{\text{ 15 min}} = \textbf{8.0 cm}^{\mathbf{3}}\cdot \textbf{min}^{\mathbf{-1}}\\\\\text{The rate of formation of oxygen is $\large \boxed{\textbf{8.0 cm}^{\mathbf{3}}\cdot \textbf{min}^{\mathbf{-1}}}$}

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Technetium-104 has a half-life of 18.0 minutes. How much of a 165 g sample is left after 90.0 minutes?
asambeis [7]

Answer:

After 90 minutes 5.15625 g amount left from total of 165 g.

Explanation:

Given data:

Technetium-104 half life = 18.0 min

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Amount left after 90.0 min = ?

Solution:

First of all we will calculate the number of half lives passes.

Number of half lives = T elapsed / Half life

Number of half lives = 90 min /18.0 min

Number of half lives =  5

Amount left:

At time 0 = 165 g

At first half life = 165 g/ 2= 82.5 g

At 2nd half life = 82.5 g/2  = 41.2 g

At 3rd half life = 41.2 g/ 2 = 20.625 g

At 4th half life =  20.625 g/ 2  = 10.3125 g

At 5th half life = 10.3125 g /2 = 5.15625 g

Thus, after 90 minutes 5.15625 g amount left from total of 165 g.

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4 years ago
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Answer/Explanation:

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