The simple formula is C = n/V
n = mols
C = Concentration or Molarity
V = Volume in Liters.
n = 2
V = 4
C = 2 / 4
C = 0.5 mol/Litre
Answer:
"The five factors are: the latitude zone, the elevation of the land, topography (landforms), ocean currents and winds."
Explanation:
Answer:
Oxalic acid is a dicarboxylic acid and forms sodium salt with NaOH and water
Explanation:
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in of ethane gas
And, as we know that
1 mole of ethane molecule contains molecules of ethane
2.869 moles of ethane molecule contains molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.