Answer:
The limiting reacting is O2
Explanation:
Step 1: data given
Number of moles O2 = 21 moles
Number of moles C6H6O = 4.0 moles
Step 2: The balanced equation
C6H6O + 7O2 → 6CO2 + 3H2O
Step 3: Calculate the limiting reactant
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
O2 is the limiting reactant. It will completely be consumed (21 moles).
C6H6O is in excess.
For 7 moles O2 we need 1 mol C6H6O
For 21 moles O2 we'll need 21/7 = 3 moles C6H6O
There will remain 4.0 - 3.0 = 1 mol C6H6O
Step 4: calculate products
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2
For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O
The limiting reacting is O2
B) the bacteria that live in the intestine of a rabbit
Answer:
How to convert volts to electron-volts
How to convert electrical voltage in volts (V) to energy in electron-volts (eV).
You can calculate electron-volts from volts and elementary charge or coulombs, but you can't convert volts to electron-volts since volt and electron-volt units represent different quantities.
Volts to eV calculation with elementary charge
The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electric charge Q in elementary charge or proton/electron charge (e):
E(eV) = V(V) × Q(e)
The elementary charge is the electric charge of 1 electron with the e symbol.
So
electronvolt = volt × elementary charge
or
eV = V × e
Example
What is the energy in electron-volts that is consumed in an electrical circuit with voltage supply of 20 volts and charge flow of 40 electron charges?
E = 20V × 40e = 800eV
Volts to eV calculation with coulombs
The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C) divided by 1.602176565×10-19:
E(eV) = V(V) × Q(C) / 1.602176565×10-19
So
electronvolt = volt × coulomb / 1.602176565×10-19
or
eV = V × C / 1.602176565×10-19
Example
What is the energy in electron-volts that is consumed in an electrical circuit with voltage supply of 20 volts and charge flow of 2 coulombs?
E = 20V × 2C / 1.602176565×10-19 = 2.4966×1020eV
Explanation:
