An empirical formula is the "reduced" version of a molecular formula. For example, CH3 is the empirical formula for C2H6, C3H9, C4H12, and so forth. The difference in subscripts between an empirical formula and molecular formula is given by the constant n. If n is a whole number, this means the numerator is the molecular formula. So the answer is D. <span>subscript of C in molecular formula = n  subscript of C in empirical formula. This can be rewritten as:
n = subscript of C in molecular formula/subscript of C in empirical formula</span>
Explanation:
I think a is correct answer
Answer:
CaCO3 (s) → CaO (s) + CO2 (g)
The mass of carbonate that must have reacted was 43.03 grams
Explanation:
CaCO3 → CaO + CO2
Relation between reactant and product is 1:1
Let's apply the Ideal Gas Law to find out the moles of CO2 which were produced.
P . V = n . R . T
1 atm . 23 L = n . 0.082 L.atm/mol.K . 653K
(1atm . 23L) / (0.082 mol.K/L.atm . 653K) = n
0.43 moles = n
0.43 moles of CO2, were produced from 0.43 moles of CaCO3.
Molar weight of CaCO3 = 100.08 g/m
Mass = Molar weight . moles
Mass = 100.08 g/m 0.43 m = 43.03 g
Answer:
Explanation:
When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acid combines with the OH(-) anion of the base to form water. The compound formed by the cation of the base and the anion of the acid is called a salt.
Explanation:
Note: Molar masses of elements can be found online or in the periodic table.
Moles of Magnesium
= 3.60g / (24.3g/mol) = 0.148mol.
Moles of Chlorine
= 10.65g / (35.45g/mol) = 0.300mol.
Mole ratio of Magnesium to Chlorine
= 0.148mol : 0.300mol = 1 : 2.
Hence we have the empirical formula MgCl2.
Moles of Lithium
= 9.1g / (6.94g/mol) = 1.311mol.
Moles of Oxygen
= 10.4g / (16g/mol) = 0.650mol.
Moles ratio of Lithium to Oxygen
= 1.311mol : 0.650mol = 2 : 1.
Hence we have the empirical formula Li2O.
