Draw the structure of the product formed by the reaction between methyl alcohol and formic acid and give the molecular formula
HCOOH + HOCH3 ==> HCOOCH3 + H2O
determine this molar concentration 58g of lithium in 54-ml of water
molarity = mols/L
mols Li = 58/atomic mass Li=??
L = 0.054.
To draw the structure of the product formed by the reaction between methyl alcohol (HOCH3) and formic acid (HCOOH), you need to understand the reaction. The reaction between these two compounds results in the formation of methyl formate (HCOOCH3) and water (H2O).
The balanced chemical equation for this reaction is:
HCOOH + HOCH3 → HCOOCH3 + H2O
The molecular formula of methyl formate is HCOOCH3.
To determine the molar concentration of 58 g of lithium in 54 mL of water, you can use the formula for molarity:
Molarity (M) = moles of solute / liters of solution
First, you need to determine the moles of lithium (Li) in the given mass.
Molar mass of Li = 6.94 g/mol
moles of Li = mass of Li / molar mass of Li = 58 g / 6.94 g/mol = 8.35 mol
Next, you need to convert the volume of water from mL to L:
Liters of water = milliliters of water / 1000 = 54 mL / 1000 = 0.054 L
Now you can calculate the molarity:
Molarity (Li) = moles of Li / liters of water = 8.35 mol / 0.054 L ≈ 154.63 M
Therefore, the molar concentration of 58 g of lithium in 54 mL of water is approximately 154.63 M.