with strong heating calcium carbonate undergoes thermal decomposition how many mole of CaCO3 are there in 50g of calcium carbona
2 answers:
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Answer:
480 L
Explanation:
In order to solve this question, you should be familiar with gas laws. (I will attach a picture showing all of them under my answer.) In this question in particular, however, we only need Charles's Law because we're dealing with temperature and volume.
As we can see, Charles's Law is:
or, initial volume over initial temperature equals final volume over final temperature.
In this question, 60 L is our <u>initial volume,</u> and 0.5 K is our <u>initial temperature</u> (K being Kelvin). We are only given 4 K as our <u>final temperature</u>. We are asked to solve for the <u>final volume</u>. Let's set up the equation and solve for :
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(60) / (0.5) = / (4)
↓
120 = / 4
×4 ×4
↓
= 480 L
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There's our answer! Feel free to comment if you have any questions about my answer :)
