This is possible because of the emulsifying properties present in soap. This property is caused by the hydrophilic end and hydrophobic end of a soap molecule. Grease is able to be dissolved in the water because it is attracted to the hydrophobic end of the soap molecule.
Answer:
115g/mol
Explanation:
To get the molar mass, we know that the it is equal to the mass divided by the number of moles. We have the mass but we do not have the number of moles.
We get this by working through the solution information. Firstly, we need to know the number of moles in 750ml for a molarity of 0.29m
Now, since 0.29 moles is present in 1000ml, x moles will be present in 750ml
The value of x is obtained as follows:
x = (750 * 0.29)/1000 = 0.2175 moles
Now since we have the number of moles, we can then obtain the molar mass.
Molar mass = mass/number of moles = 25.0g/0.2175 = 114.94 approximately 105g/mol
I think this one is the organelles.
Sodium is a member of the alkali metal family with potassium (K) and Lithium (LI) sodium's big claim to fame is that it's one or two elements in your table salt. when bonded to chlorine (CI) THE two elements make sodium chloride
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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