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Rus_ich [418]
3 years ago
15

How many liters of oxygen gas, at standard temperature and pressure, will react with 25.0 grams of magnesium metal? Show all of

the work used to solve this problem.
2 Mg + O2 ----> 2 MgO
Chemistry
2 answers:
love history [14]3 years ago
8 0
Given:
2 Mg + O2 → 2 MgO 
So,
<span>(25.0 g of Mg / 24.3051 g/mole) x (1/2) x (22.4 L/mole) = 11.5 L  of Oxygen will react with 25 grams of magnesium metal.</span>
qwelly [4]3 years ago
5 0

Answer : The volume of oxygen gas is, 11.67 liters

Explanation : Given,

Mass of Mg = 25 g

Molar mass of Mg = 24 g/mole

First we have to calculate the moles of Mg.

\text{Moles of }Mg=\frac{\text{Mass of }Mg}{\text{Molar mass of }Mg}=\frac{25g}{24g/mole}=1.042moles

Now we have to calculate the moles of oxygen gas.

The balanced chemical reaction is,

2Mg+O_2\rightarrow 2MgO

From the balanced chemical reaction, we conclude that

As, 2 mole of Mg react with 1 mole of oxygen gas

So, 1.042 mole of Mg react with \frac{1.042}{2}=0.521 mole of oxygen gas

Now we have to calculate the volume of oxygen gas.

As we know that at STP,

1 mole of oxygen gas contains 22.4 L volume of oxygen gas

So, 0.521 mole of oxygen gas contains 0.521\times 22.4=11.67L volume of oxygen gas

Therefore, the volume of oxygen gas is, 11.67 liters

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The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
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