How many grams of carbon dioxide are created from the complete
combustion of 21.3 L of butane at STP?*

167.2 g

Explanation:

Known

VC4H10 = 21.3

T = 0.00 C (convert to Kelvin: 273 K)

P = 1.00 atm

Unknown

m = ?g

1. Write the balanced chemical equation

1 C4H10 + 1O2  >  4 CO2 + 5 H2O

2. Find the volume ratio of Carbon Dioxide to Butane

1 C4H10   4 CO2  = 4 volumes CO2 / 1 volume C4H10

3. Multiply by the known volume of n (butane)

21.3 L C4H10  x  4 volumes CO2 / 1 volume C4H10 = 85.2 L C4H10

4. Use ideal gas law

PV = nRT   solve for n  > n = PV/RT

n= (1.00 atm) (85.2 L) / (0.0821 L atm/mol K) (273)  = 3.80 mol CO2

5. Find molar mass of CO2

1 C x 12 + 2 O x 16 = 44.00

6. Multiply the ideal gas law solution (3.80) by molar mass CO2 (44.00)

3.80 mol CO2 x 44.00 g CO2

= 167.2 g CO2  

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current