The type of the bond is present Na₃PO₄ is the ionic bond. the Na₃PO₄ is the ionic compound. yes the Na₃PO₄ is the polyatomic ion.
The Na₃PO₄ is Na⁺ and PO₄³⁻. the phosphorus is the non metal and the oxygen atom is the non metal. the non meta and non meta form the covalent or molecular bond. the bond between the PO₄³⁻ bond is the covalent bond but the overall present in the Na₃PO₄ is the ionic bond . the bons in between the Na⁺ and PO₄³⁻ is the the ionic bond. the PO₄³⁻ id the polyatomic ion .
The bond between the positively charged ion and the negatively charged ion are called as the ionic bond and the compound form is the ionic compound.
To learn more about ionic bond here
brainly.com/question/29005103
#SPJ4
Answer:
babies have the most sensitive taste buds
Answer:
the atomic mass of any elemet contains avogardo numberof atoms
In case of Gallium,
69.72 gram is atomic mass and it cotnains around 6.023*10^23 atoms of Gallium
but, 2000 punds = 907184.7 grams
907184.7 gram of gallium contains= 6.023*10^23* 907184/69.72
= 79 *10^26 atoms
Explanation:
Answer: For the elementary reaction
the molecularity of the reaction is 2, and the rate law is rate = ![k[NO_3]^1[CO]^1](https://tex.z-dn.net/?f=k%5BNO_3%5D%5E1%5BCO%5D%5E1)
Explanation:
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.
Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction. Thus it can never be fractional.
For elementary reaction
, molecularity is 2 and rate law is ![rate=k[NO_3]^1[CO]^1](https://tex.z-dn.net/?f=rate%3Dk%5BNO_3%5D%5E1%5BCO%5D%5E1)