Find the empirical formula of each of the following compounds. given mass or for each element in a sample of the compound 3,611 g ca; 6.389 g c1

CaCl₂  

Step-by-step explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Ca to Cl.

Data:

Mass of Ca = 3.611 g

Mass of Cl = 6.389 g

Calculations

Step 1. Calculate the moles of each element

Moles of Ca = 3.611 g Ca × (1 mol Ca/(40.08 g Ca)= 0.090 10 mol Ca

Moles of Cl = 6.389 g Cl  

Step 2. Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

Ca:Cl = 0.090 10:0.1802 = 1:2.000

Step 3. Round the molar ratios to the nearest integer

Ca:Cl = 1:2.000 ≈ 1:2

Step 4: Write the empirical formula

EF = CaCl₂

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.