Based on the following reaction, determine how much heat is released if 500. grams of methane is combusted in excess oxygen: CH4 + 2 O2  CO2 + 2 H2O H= – 890 kJ

- 13,150.6kJ

Explanation:

CH4 + 2 O2 > CO2 + 2 H2O ΔH= – 890 kJ

The ΔH is enthalpy change of combustion , which is the heat is either absorbed or released by the combustion of one mole of a substance.

ΔH=−890 kJ/mol (released in the combustion of one mole of methane)

using the molar mass  (in grams )of methane to get moles of sample

(237g × 1 mole of CH4)/16.04g=14.776 moles  of CH4

Since 1 mole produces 890 kJ of heat upon combustion, then 14.776 moles will produce

ΔH = 14.776moles of CH4 × 890kJ/1mole of CH4

=13,150.6kJ

Therefore ΔH = - 13,150.6kJ

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

metric time was an attempt to create a standard unit of time based off of 10, but it never took off an is not included in the metric system (si)