Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:

a) moles of butane

b) moles of oxygen


According to stoichiometry :
2 moles of butane require 13 moles of 
Thus 0.09 moles of butane will require =
of 
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.
C = 0.11 mol
V = 5.65 L
n = ???
n = C*V
n = 0.11 * 5.65
n = 0.622 mols
1 mol of CaCl2 = 40 + 2*35.5 = 111 grams
0.622 mol = x
x = 111 * 0.622
x = 69.0 grams CaCl2
Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J
Explanation:
1) Data:
Water ⇒ C = 1 cal/g°C
m = 65.8 g
Ti = 31.5°C
Tf = 36.9°C
Heat, Q = ?
2) Formula:
Q = mCΔT
3) Calculations:
Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal
4) You can convert from calories to Joules using the conversion factor:
1 cal = 4.18 J
⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J
Answer:
The Ideal gas law
Explanation:
From the given question, we have:
V

where each variable has its usual meaning.
Thus,
V = 
where R is the ideal gas constant
cross multiply to have;
PV = nRT
This implies that the volume of the gas is directly proportional to the number of moles of the gas.
Therefore, the law can be used to determine the relationship between the volume and number of moles is the ideal gas law.