At what temperature (inc) would the volume a gas be equal to 45.7L if the volume of gas was 33.9L at 12.4c

The answer to your question is  T1 = 384.7 °K

Explanation:

Data

Volume 1 = V1 = 45.7 l

Temperature 1 = T1 = ?

Volume 2 = V2 = 33.9 l

Temperature 2 = T2 = 12.4°C

To solve this problem use Charles' law

              V1/T1 = V2/T2

                    T1 = V1T2/V2

-Convert temperature to °K

T2 = 12.4 + 273 = 285.4°K

-Substitution

                    T1 = (45.7 x 285.4) / 33.9

-Simplification

                    T1 = 13042.8 / 33.9

-Result

                    T1 = 384.7 °K

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"