4.2g of cerium reacted with oxygen to form 5.16g of an oxide of cerium. Find the simplest formula of this oxide*
ASAP plz

CeO₂

Explanation:

Hello!

In this case, since we are given the mass of both cerium and the cerium oxide, we can first compute the moles of cerium and the moles of oxygen as shown below:

Now, we simply divide each moles by 0.03 as the fewest moles in the formula to obtain the simplest formula (empirical formula) of this oxide:

Thus, the formula turns out:

Regards!

d. temperate deciduous forest

"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"