To answer this question, we will use the following equation:
<span>ln(P2/P1) = (∆Hvap/R)*((1/T1) - (1/T2))
</span>
Now we examine the givens of the problem and transform to standard units if required:
<span>∆Hvap = 30.5 kJ/mol
</span>R is a constant = <span>8.314 x 10^-3 kJ K^-1 mol^-1
T1 </span><span>= 91 celcius = 91 + 273= 364 Kelvin
</span>T2 = 20 celcius = 20 + 273 = 293 k3lvin
P1 is the standard atmospheric pressure = 760 mmHg
P2 is the value to be calculated
Substitute with these values in the equation:
ln(P2/760) = (30.5 / 8.314 x 10^-3) x ((1 / 364) - (1 / 293))
ln(P2/760) = - 2.4662 (Take the exponential both sides to eliminate the ln)
P2 / 760 = e^(-2.4462) = 0.0866
P2 = 0.0866 x 760 = 65.816 mmHg
Answer:
709 g
Step-by-step explanation:
a) Balanced equation
Normally, we would need a balanced chemical equation.
However, we can get by with a partial equation, as log as carbon atoms are balanced.
We know we will need an equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 30.07 236.74
C₂H₆ + … ⟶ C₂Cl₆ + …
m/g: 90.0
(i) Calculate the moles of C₂H₆
n = 90.0 g C₂H₆ × (1 mol C₂H₆ /30.07 g C₂H₆)
= 2.993 mol C₂H₆
(ii) Calculate the moles of C₂Cl₆
The molar ratio is (1 mol C₂Cl₆/1 mol C₂H₆)
n = 2.993 mol C₂H₆ × (1 mol C₂Cl₆/1 mol C₂H₆)
= 2.993 mol C₂Cl₆
(iii) Calculate the mass of C₂Cl₆
m = 2.993 mol C₂Cl₆ × (236.74 g C₂Cl₆/1 mol C₂Cl₆)
m = 709 g C₂Cl₆
The reaction produces 709 g C₂Cl₆.
Answer:
The amount of sodium in this sample = 32.4 grams
Explanation:
Mass of Sodium = 23 g/mol
Mass of Chlorine = 35.5 g/mol
Mass of NaCl = 35.5 g/mol
Fraction of Na in NaCl = 
=0.648
We are given a sample of 50 grams of sodium chloride.
The amount of sodium in this sample = 
The amount of sodium in this sample = 
The amount of sodium in this sample = 32.4 grams
