Answer:
th mol
Explanation:
When you balance an equation, you can refer back to the coefficients when calcuating mol and molar mass from the given information about one reactant to antoher reactant ot product. you use the coeffiecients for mol-mol ratio.
for example. if you are given A +3 B -----> 6C + 5D, and you have 10 mol of A and you have to find how many mols of C with 10 mol of A. then you would ratio it as A: C = 10A: 10C= 10(1): 10(6)= 10:60. so with 10 mol A you can make 60 mol C.
Answer:
An emulsifying agent is typically characterized by having <u><em>d. one polar end and one nonpolar end.</em></u>
Explanation:
Emulsifiers are substances that have the ability to bind, for example, fats with those substances that have mostly water in their conformation. In other words, the emulsifier facilitates mixtures of two or more immiscible liquid substances.
This is because the molecules of an emulsifier are often lipophilic (attract oil) at one end and hydrophilic (attract water) at the other. In other words it consists of a polar (hydrophilic) head group and a non-polar (hydrophobic) tail.
<u><em>An emulsifying agent is typically characterized by having d. one polar end and one nonpolar end.</em></u>
Answer:
Boiling point: 63.3°C
Freezing point: -66.2°C.
Explanation:
The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:
<em>Boiling point increasing:</em>
ΔT = kb*m*i
<em>Freezing point depression:</em>
ΔT = kf*m*i
ΔT are the °C that change boiling or freezing point.
m is molality of the solution (moles / kg)
And i is Van't Hoff factor (1 for I₂ in chloroform)
Molality of 50.3g of I₂ in 350g of chloroform is:
50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:
0.198 moles / 0.350kg = 0.566m
Replacing:
<em>Boiling point:</em>
ΔT = kb*m*i
ΔT = 3.63°C/m*0.566m*1
ΔT = 2.1°C
As boiling point of pure substance is 61.2°C, boiling point of the solution is:
61.2°C + 2.1°C = 63.3°C
<em>Freezing point:</em>
ΔT = kf*m*i
ΔT = 4.70°C/m*0.566m*1
ΔT = 2.7°C
As freezing point is -63.5°C, the freezing point of the solution is:
-63.5°C - 2.7°C = -66.2°C
Using the Equation: PV=nRT
Where P is the pressure 60 cmHg or 600 mmHg or 600/760= 0.789 atm
V is the volume 125 ml or 0.125 L, n is the number of moles, R is a constant 0.082057, and T is temperature 25 °C or 298 K;
Therefore:
0.789 × 0.125 = n × 0.082057 × 298
n = 0.0987/24.45
= 0.004036 mol
0.004036 mole has a mass of 0.286 g
Hence; 1 mole has a mass of 0.286/0.004036
= 70.8 g /mol
Therefore the molar mass of the gas is 71 g/mol (2 sfg)
<u>Answer:</u> The standard Gibbs free energy of the given reaction is 6.84 kJ
<u>Explanation:</u>
For the given chemical equation:
The expression of 
for above equation follows:
We are given:
Putting values in above expression, we get:
To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:
where,
= standard Gibbs free energy = ?
R = Gas constant = 8.314 J/K mol
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
= equilibrium constant at 25°C = 0.0632
Putting values in above equation, we get:
Hence, the standard Gibbs free energy of the given reaction is 6.84 kJ
