Here is the complete question.
Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissolves readily in water and polar organic solvents like ethanol. Calculate the mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol (CH3CH2OH; density = 0.7893 g/mol). Round to four significant digits
Answer:
0.9567 mol
Explanation:
Given that:
mass of glycerol = 1.61 g
molar mass of glycerol = 92.1 g/mol
no of mole =
∴ number of moles of glycerol () =
= 0.0175 mol
Volume of ethanol = 22.60 mL
Density of ethanol = 0.7893 g/mL
Since Density =
∴ mass of ethanol = density of ethanol × volume of ethanol
mass of ethanol = 0.7893 g/mL × 22.60 mL
mass of ethanol = 17.838 g
Number of moles of ethanol =
= 0.387 mole
∴ the mole fraction of the solvent can be determined as:
= 0.95673671199
≅ 0.9567 mol
∴ The mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol is = 0.9567 mol
<em>The statement that gives the relationship between energy needed in breaking a bond and the one that is released after breakin</em>g is
The amount of energy it takes to break a bond is always less than the amount of energy released when the bond is formed.
- Bond energy can be regarded as amount of energy that is required in breaking a particular bond.
- For a bond to be broken Energy will be added and when a bond is broken there will be release of energy
- Bond breaking can be regarded as endothermic process, it is regarded as endothermic because there is a lot of energy required to be absorbed.
- Where ever a bond is broken, there must be formation of another bond
- Bond forming on the other hand can be regarded as exothermic process, since there is a release of releases energy.
Therefore, more energy is required in breaking of bond compare to energy released after breaking of bond.
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Answer:
Option A = 2.2 L
Explanation:
Given data:
volume of one mole of gas = 22.4 L
Volume of 0.1 mole of gas at same condition = ?
Solution:
It is known that one mole of gas at STP occupy 22.4 L volume. The standard temperature is 273.15 K and standard pressure is 1 atm.
For 0.1 mole of methane.
0.1/1 × 22.4 = 2.24 L
0.1 mole of methane occupy 2.24 L volume.
Answer:
The total weight of both object is 78.56 kg.
Explanation:
Given data:
Mass of object A = 45.1 kg
Mass of object B = 33.46 kg
Total weight of object = ?
Solution:
Total weight of both subject must be the sum of weight of object A and B.
Total weight of objects = weight of object A + weight of object B
Now we will put the values of mass of object A and B.
Total weight of objects = 45.1 kg + 33.46 kg
Total weight of objects =78.56 kg
Thus the total weight of both object is 78.56 kg.