If 14.5 grams of silver nitrate (molar mass =169.88g/mol) reacts with aluminum chloride. How many grams of aluminum nitrate (mol
1 answer:
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Answer:
44.2 L
Explanation:
Use Charles Law:
We have all the values except for V₂; this is what we're solving for. Input the values:
- make sure that your temperature is in Kelvin
From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:
Therefore, V₂ = 44.2 L
It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.
Answer: The volume of the sample after the reaction takes place is 29.25 L.
Explanation:
The given reaction equation is as follows.
So, moles of product formed are calculated as follows.
Hence, the given data is as follows.
= 0.17 mol,
= 0.255 mol
= 19.5 L,
As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.