https://openstax.org/books/chemistry-2e/pages/3-2-determining-empirical-and-molecular-formulas
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Answer : The molarity of solution is, 1.73 mole/L
Explanation :
The relation between the molarity, molality and the density of the solution is,
where,
![d=M[\frac{1}{m}+\frac{M_b}{1000}]](https://tex.z-dn.net/?f=d%3DM%5B%5Cfrac%7B1%7D%7Bm%7D%2B%5Cfrac%7BM_b%7D%7B1000%7D%5D)
d = density of solution = 
m = molality of solution = 2.41 mol/kg
M = molarity of solution = ?
= molar mass of solute (toluene) = 92 g/mole
Now put all the given values in the above formula, we get the molality of the solution.
![0.876g/ml=M\times [\frac{1}{2.41mol/kg}+\frac{92g/mole}{1000}]](https://tex.z-dn.net/?f=0.876g%2Fml%3DM%5Ctimes%20%5B%5Cfrac%7B1%7D%7B2.41mol%2Fkg%7D%2B%5Cfrac%7B92g%2Fmole%7D%7B1000%7D%5D)
Therefore, the molarity of solution is, 1.73 mole/L
Answer: Hydrogen bonds
Explanation: Hydrogen bonds allow two molecules to link together temporarily. Water molecules are made up of two hydrogen atoms and one oxygen atom, held together by polar covalent bonds.
Answer:
Theoretical yield: 13.9 g
Percent yield: 94 %
Explanation:
The reaction is:
CaCO₃ → CaO + CO₂
We see that the reaction is correctly balanced.
1 mol of calcium carbonate can decompose to 1 mol of calcium oxide and 1 mol of carbon dioxide.
We convert the mass to moles: 24.8 g . 1mol / 100.08g = 0.248 moles
As ratio is 1:1, 0.248 moles of salt can decompose to 1 mol of oxide.
We convert the moles to mass: 0.248 mol . 56.08g /1mol = 13.9 g
That's the theoretical yield.
To determine the percent yield we think:
(Determined yield / Theoretical yield) . 100 → (13.1 / 13.9) . 100 = 94 %
