Answer:
20.4g
Explanation:
First, we need to write a balanced equation for the reaction of butane to produce carbon dioxide.
When butane undergo Combustion, it will produce carbon dioxide and water as shown below:
2C4H10 + 13O2 —> 8CO2 + 10H2O
Molar Mass of C4H10 = (12x4) + (10x1) = 48 + 10 = 58g/mol
Mass of C4H10 from the balanced equation = 2 x 58 = 116g
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 8 x 44 = 352g
From the equation,
116g of C4H10 produced 352g of CO2.
Therefore, Xg of C4H10 will produce 61.9g of CO2 i.e
Xg of C4H10 = (61.9 x 116)/352 = 20.4g
Therefore, 20.4g of butane(C4H10) is needed to produce 61.9g of CO2
Molar solubility is number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturated.
The molar solubility of lead(ii) chloride with ksp value of 2.4 × 10e4 can be solve as:
Ksp = s2 = 2.4 × 10e4
s2 = 2.4 × 10e4
s = √(2.4 × 10e4)
s = 154.9 mol/L
Answer:
Explanation:
A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters
PV =nRT
at STP P= 1atm. T= 273 K
n is the number of moles. O2 has a molar mass of 32.
30 gm of O2 is 30/32= 0.94 =n
PV = nRT
at STP: P= 1 atm, T=273 K, R is always 0.082 for P in atm and T in K
SO
1 X V = 0.94 X 0.082 X 273
using high school freshman algebra,
V= 0.94 X 0.082 X 273 = 21L
using high school algebra I,
V=
Electrons are added to the same principal energy level.
