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Archy [21]
3 years ago
14

Aq. sodium oxalate reacts with aq calcium chioride to form sd nd soum ce

Chemistry
1 answer:
STatiana [176]3 years ago
4 0
Na2C2O4(aq) + CaCl2(aq) -----> 2NaCl(aq) + CaC2O4(s)

Here, CaC2O4(s) is a precipitate in the reaction as a result of precipitation reaction or double displacement reaction.

As we know that double displacement reaction two metal ions displaces each other from their salt solutions.

As we know that precipitation reaction is a reaction in which precipitate is formed.
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Why does on emole of carbon have a smaller mass than one mole of sulfur? how are the atomic structures of these elements differe
Setler79 [48]
Because carbon has a molar mads of 12 (6 protons and 6 neutrons) while sulfur has a molar mass of 16 (8 protons and 8 neutrons)
8 0
3 years ago
What is point group of allene?​
Fantom [35]

Allene (1,2-propadiene) has point group D2d, itself is achiral because it has two planes of symmetry. ... An allene with substituents on one terminal carbon atom are unlike and substituent on other terminal carbon atoms are same, allene will be achiral. It will have one symmetry plane.

Hope this helped :)

4 0
4 years ago
Read 2 more answers
In a voltaic cell, the ____
alekssr [168]
In a voltaic cell , the anode electrons and is oxidized , while the cathode electrons and is reduced.
5 0
3 years ago
Read 2 more answers
When 2 grams of powdered lead (IV) oxide was added to 100 cm3 of hydrogen peroxide, water and oxygen were produced. Lead (IV) ox
kirza4 [7]

Answer:

The right answer is:

Replacing the powdered lead oxide with its large crystals

Removing lead (IV) oxide from the reaction mixture

Using 1.0 gram of lead (IV) oxide

Explanation:

Based on the given information this reaction is the catalytic decomposition of H₂O₂ into water and oxygen  using Lead (IV) oxide as a catalyst.

  1. The catalyst surface area is directly proportional to the reaction rate
  • So, Replacing the powdered lead oxide with its large crystals would decrease the reaction rate due to the has larger surface area than its large crystals.

     2. Also, Removing lead (IV) oxide from the reaction mixture  the reaction rate decreased because as the catalyst is removed.

     3.  Using 50 cm³ of hydrogen peroxide  doesn't affect the rate because the concentration of the reactant doesn't change.

     4. Using 1.0 gram of lead (IV) oxide would decrease the reaction rate because the amount of catalyst decreased

So, The right answer is:

Replacing the powdered lead oxide with its large crystals

Removing lead (IV) oxide from the reaction mixture

Using 1.0 gram of lead (IV) oxide

7 0
4 years ago
Instead of 6 M NaOH being added to the solution, 6 M HCl is added. How will this affect the test for the presence of ammonium io
Lera25 [3.4K]

Answer:

Ammonia gas(an alkaline gas with characteristics of choking or irritating smell) is not liberated when 6mole of HCl is added to the solution instead of 6mole of NaOH, to test for the presence of ammonium ion in the solution

Explanation:

As expected, when testing for ammonium ion in a solution (precisely ammonium salt solution), Sodium Hydroxide (NaOH) is required as the test reagent.

When NaOH is added to the solution, A gas with characteristics of choking or irritating smell is liberated.

This gas turn red litmus paper blue.

This liberated gas is an alkaline gas, which is confirmed as an ammonia gas(NH3).

If HCl is added instead of NaOH, the ammonia gas will not be liberated, which indicates that the test reagent used is wrong.

3 0
3 years ago
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