Greetings!
To find the empirical formula you need the relative atomic mass of each element!
Li = 6.9
C = 12
O = 16
You can simply change the percentages into full grams
Li = 18.8g
C = 16.3g
O = 64.9
Then you use this to find the Number of moles = amount in grams / atomic mass
Li = 18.8 ÷ 6.9 = 2.7246
C = 16.3 ÷ 12 = 1.3583
O = 64.9 ÷ 16 = 4.0562
Then divide each number of moles by the smallest value:
Li = 2.7246 ÷ 1.3583 = 2.0
C = 1.3583 ÷ 1.3583 = 1
O = 4.0562 ÷ 1.3583 = 2.9 ≈ 3
So that means that there are 2 Li, 1 C, and 3 O
Empirical formula would be:
Li₂CO₃
Hope this helps!
<h2>Answer:</h2>
Option (B):
The products can form reactants, and the reactants can form products.
<h3>Explanation:</h3><h3>Reversible reaction</h3>
A reversible reaction is a reaction where the reactants form products, which react together to give the reactants back.
aA + bB ⇄ cC + dD
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B.
Other options are wrong because off:
(A) Concentration changes with time equilibrium concentration and higher product concentration is also possible.
(C) They may be constant.
(D) Concentration changes with time equilibrium concentration and higher reactant concentration is also possible.
Answer:
There are three main sources of heat in the deep earth: (1) heat from when the planet formed and accreted, which has not yet been lost; (2) frictional heating, caused by denser core material sinking to the center of the planet; and (3) heat from the decay of radioactive elements.
Moles of phosphoric acid would be needed : 0.833
<h3>Further explanation</h3>
Given
15 grams of water
Required
moles of phosphoric acid
Solution
Reaction(decomposition) :
H3PO4 -> H2O + HPO3
mol water (H2O :
= mass : MW
= 15 g : 18 g/mol
= 0.833
From the equation, mol ratio H3PO4 = mol H2O = 1 : 1, so mol H3PO4 = 0.833
Its phosphorus (P)In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.
