HCOOH is an acid.
Sr(HCOO)2.
Ca(HCOO)2
Formic acid has the formula HCOOH and is a liquid. When placed in water it forms an aqueous solution and it partially ionizes to H^+(aq) + HCOO^-(aq) or more correctly it ionizes in H2O as follows: HCOOH + H2O ==> HCOO^- + H3O^+
HCHO2 is a weird way of writing formic acid, which is usually written CHOOH. As you might be able to guess from it's name, it's an acid. KOH is a base. Like most acids and bases, they react to make water, and a salt. The salt in this case would be potassium formate.
neither, Formic acid does not completely dissociate in water so it is a weak acid. HCOOH + H2O <=> HCOO- + H3O+
yes, in the balanced molecular equation: 2Al(s)+6HCOOH(aq)->2Al(HCOO)3(aq)+3H2(g)
Calcium carbonate + methanoic acid ---> calcium methanoate + carbon dioxide + waterCaCO3 + 2HCOOH ---> Ca2+(HCOO-)2 + CO2 + H2O
The conjugate base of perchloric acid is the ion chlorate (ClO4)-.
HCHO2 is methanoic acid and is normally written as HCOOH. When reacted with potassium hydroxide (KOH) , it produces potassium methanoate. and water/ HCOOH +KOH = HCOO^-K+ + H2O
When any carboxylic acid is place in water, hydrogen ion transfer occurs to produce hydronium ion and carboxylate ion. ( R-COOH + H20 = H30 + R-COO- ) So formic acid with the addition of water is HCOOH + H20 = H30 + HCOO-
Chloral does not undergo Cannizzaro reaction because the anion formed is very stable do to the electron withdrawing effect of the three chloro group attached to the alpha carbon. As a result cleavage takes place and the products formed initially are CCl3- and HCOOH. Then by maintaining Bronsted Lowry concept an hydride ion transfer takes place and the final products are chloroform and HCOO- ion.