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.
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Answer:
There is 2.52 kJ of energy released (option 4)
Explanation:
Step 1: Data given
The enthalpy of fusion of methanol (CH3OH) is 3.16 kJ/mol
Mass of methanol = 25.6 grams
Molar mass of methanol = 32.04 g/mol
Step 2: Calculate moles of methanol
Moles methanol = mass methanol / molar mass methanol
Moles methanol = 25.6 grams / 32.04 g/mol
Moles methanol = 0.799 moles
Step 3: Calculate energy transfer
Energy transfer = moles * enthalpy of fusion
Energy = 0.799 moles * 3.16 kJ/mol
Energy = 2.52 kJ released
There is 2.52 kJ of energy released
THE ANSWER IS: <u>737.5</u>
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Answer:
what are the orbitals present in the fifth principal energy level
Explanation:
Answer : The concentration of HI (g) at equilibrium is, 0.643 M
Explanation :
The given chemical reaction is:
Initial conc. 0.10 0.10 0.50
At eqm. (0.10-x) (0.10-x) (0.50+2x)
As we are given:
The expression for equilibrium constant is:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
Now put all the given values in this expression, we get:
x = 0.0713 and x = 0.134
We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.0713
The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M
Thus, the concentration of HI (g) at equilibrium is, 0.643 M
