The full ionic equation for the conversion of aqueous lead (II) nitrate to solid lead (II) sulfate and aqueous potassium nitrate is as follows:
<h3>What is
ionic equation ?</h3>
PbSO4(s) + 2 K+ + 2 NO3 = Pb2+(aq) + 2 NO3(aq), 2 K+(aq), and SO42(aq) (aq)
Let's look at the molecular formula that results in solid lead (II) sulfate and aqueous potassium nitrate when aqueous lead (II) nitrate and aqueous potassium sulfate interact. This reaction involves two displacements.
PbSO4(s) + 2 KNO3 = Pb(NO3)2(aq) + K2SO4(aq) (aq)
All ions and molecular species are included in the full ionic equation.
PbSO4(s) + 2 K+ + 2 NO3 = Pb2+(aq) + 2 NO3(aq), 2 K+(aq), and SO42(aq) (aq)
The full ionic equation for the conversion of aqueous lead (II) nitrate to solid lead (II) sulfate and aqueous potassium nitrate is as follows:
PbSO4(s) + 2 K+ + 2 NO3 = Pb2+(aq) + 2 NO3(aq), 2 K+(aq), and SO42(aq) (aq)
To learn more about ionic equation refer to
brainly.com/question/13879496
#SPJ4
Mass defect and binding energy are related as
ΔE = Δmc^2
Where
ΔE = binding energy
Δm = mass defect
c = speed of light
given
mass defect = 3.09x 10-27 kg
We know that speed of light = 3 X 10^8 m /s
ΔE = 3.09x 10-27 kg (3 X 10^8 m /s)^2 = 2.781 X 10^-10 J / Kg
.
.
Answer : The heat of the reaction is, 1.27 kJ/mole
Explanation :
First we have to calculate the heat released.
Formula used :
![Q=m\times c\times \Delta T](https://tex.z-dn.net/?f=Q%3Dm%5Ctimes%20c%5Ctimes%20%5CDelta%20T)
or,
![Q=m\times c\times (T_2-T_1)](https://tex.z-dn.net/?f=Q%3Dm%5Ctimes%20c%5Ctimes%20%28T_2-T_1%29)
where,
Q = heat = ?
m = mass of sample = 1.50 g
c = specific heat of water = ![4.81J/g^oC](https://tex.z-dn.net/?f=4.81J%2Fg%5EoC)
= initial temperature = ![22.7^oC](https://tex.z-dn.net/?f=22.7%5EoC)
= final temperature = ![19.4^oC](https://tex.z-dn.net/?f=19.4%5EoC)
Now put all the given value in the above formula, we get:
![Q=1.50g\times 4.81J/g^oC\times (19.4-22.7)^oC](https://tex.z-dn.net/?f=Q%3D1.50g%5Ctimes%204.81J%2Fg%5EoC%5Ctimes%20%2819.4-22.7%29%5EoC)
![Q=-23.8095J=-0.0238kJ](https://tex.z-dn.net/?f=Q%3D-23.8095J%3D-0.0238kJ)
Now we have to calculate the heat of the reaction in kJ/mol.
![\Delta H=\frac{Q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7BQ%7D%7Bn%7D)
where,
= enthalpy change = ?
Q = heat released = 0.0238 kJ
n = number of moles NH₄NO₃ = ![\frac{\text{Mass of }NH_4NO_3}{\text{Molar mass of }NH_4NO_3}=\frac{1.50g}{80g/mol}=0.01875mole](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BMass%20of%20%7DNH_4NO_3%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DNH_4NO_3%7D%3D%5Cfrac%7B1.50g%7D%7B80g%2Fmol%7D%3D0.01875mole)
![\Delta H=\frac{0.0238kJ}{0.01875mole}=1.27kJ/mole](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B0.0238kJ%7D%7B0.01875mole%7D%3D1.27kJ%2Fmole)
Therefore, the heat of the reaction is, 1.27 kJ/mole
Answer:
The answer is in the problem
Explanation:
As general rule of number of oxygen is -2:
O → -2
Alkali metals (Li, Na, K) are always +1
Na → +1
Alkali earth methals (Be, Mg, Ca...) are always +2
Ca → +2
The halogen group (F, Cl, Br...) is always -1
F → -1
The oxidation number of Si (+/- 4)
Aluminium is, usually +3
And to complete the octet rule in nitrogen, 3 electrons are required. That means:
N → -3
Answer:
wheres the question or picture
Explanation: