The empirical formula : CH₃
<h3>Further explanation</h3>
Given
2.5 g sample
2.002 g Carbon
Required
The empirical formula
Solution
Mass of Hydrogen :
= 2.5 - 2.002
= 0.498
Mol ratio C : H :
C : 2.002/12 = 0.167
H : 0.498/1 = 0.498
Divide by 0.167 :
C : H = 1 : 3
Molality
is one way of expressing concentration of a solute in a solution. It is expressed
as the mole of solute per kilogram of the solvent. To calculate for the
molality of the given solution, we need to convert the mass of solute into
moles and divide it to the mass of the solvent.
<span>
Moles of HCl = 5.5 g HCl ( 1 mol HCl / 36.46 g HCl ) = 0.1509 mol HCl</span>
<span>
Molality = 0.1509 mol HCl / 200 g C2H6O ( 1 kg / 1000 g )
Molality
= 0.7543 mol / kg</span>
<span>The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is
0.7453 molal.</span>
Only one atom because atom with 17 electrons need that one electron to become stable
Answer:
2 sig figs.
Explanation:
Sig Fig Rules:
Any non-zero digit is a significant figure.
Any zeros between 2 non-zero digits are significant figures.
Trailing zeros after the decimal are significant figures.
