A)1.75×3 moles of carbon monoxide
B)2:3
A)each mole of ferric oxide requires 3 moles of carbon monoxide. Therefore 1.75 moles requires 1.75 ×3 moles of carbon monoxide
Volume fraction = volume of the element / volume of the alloy
Volume = density * mass
Base: 100 grams of alloy
mass of tin = 15 grams
mass of lead = 85 grams
volume = mass / density
Volume of tin = 15g / 7.29 g/cm^3 = 2.06 cm^3
Volume of lead = 85 g / 11.27 g/cm^3 = 7.54 cm^3
Volume fraction of tin = 2.06 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.215
Volume fraction of lead = 7.54 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.785
As you can verify the sum of the two volume fractions equals 1: 0.215 + 0.785 = 1.000
Answer:
0.681 atm
Explanation:
To solve this problem, we make use of the General gas equation.
Given:
P1 = 785 torr
V1 = 2L
T1 = 37= 37 + 273.15 = 310.15K
P2 = ?
V2 = 3.24L
T2 = 58 = 58+273.15 = 331.15K
P1V1/T1 = P2V2/T2
Now, making P2 the subject of the formula,
P2 = P1V1T2/T1V2
P2 = [785 * 2 * 331.15]/[310.15 * 3.24]
P2 = 515.715 Torr
We convert this to atm: 1 torr = 0.00132 atm
515.715 Torr = 515.715 * 0.00132 = 0.681 atm
