Answer:
Explanation:
mass of the solution m = 1.6 + 75 = 76.6 g
fall in temperature = 25 - 23.34 = 1.66°C
heat absorbed = mass x specific heat x fall in temperature
= 76.6 x 1.66 x 4.18
= 531.5 J .
= .5315 kJ .
mol weight of ammonium nitrate = 80 g
heat absorbed by 1.6 g = .5315 kJ
heat absorbed by 80 g or one mole = 26.575 kJ
enthalpy change ΔH = +26.575 kJ
b )
enthalpy of hydration = 2630 kJ / mol
lattice energy = enthalpy of hydration + enthalpy change
= 2630 + 26.575
= 2656.575 kJ .
Just use q=mCDeltaT
q=energy
m=mass
c=specific heat
Delta T= Change in temperature
Mercury atoms have 80 electrons and 80 protons with 122 neutrons in the most abundant isotope. Under standard conditions mercury is a shiny, heavy, silvery liquid. It is the only metal that is liquid at room temper
Answer : The standard cell potential of the reaction is, -1.46 V
Explanation :
The given balanced cell reaction is,
Here, chromium (Cr) undergoes oxidation by loss of electrons and act as an anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
The standard values of cell potentials are:
Standard reduction potential of lead ![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
Standard reduction potential of chromium ![E^0_{[Cr^{3+}/Cr]}=1.33V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D%3D1.33V)
Now we have to calculate the standard cell potential for the following reaction.
![E^0=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)
Therefore, the standard cell potential of the reaction is, -1.46 V
sp3d2 hybridization has 1s, 3p and 2d orbitals, that undergo intermixing to form 6 identical sp3d2 hybrid orbitals. These 6 orbitals are directed towards the corners of an octahedron. They are inclined at an angle of 90 degrees to one another.
<h3>Which orbitals hybridize to form hybrid orbitals?</h3>
For each carbon, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals. These hybridized orbitals align themselves in the trigonal planar structure. For each carbon, two of these sp orbitals bond with two 1s hydrogen orbitals through s-sp orbital overlap.
<h3 /><h3>What is sp3d2 hybridization?</h3>
Intermixing of one 's', three 'p' and two 'd' orbitals of almost same energy by giving six identical and degenerate hybrid orbitals is called sp3d2 hybridization. These six sp3d2 orbitals are arranged in octahedral symmetry by making 90° angles to each other.
Learn more about hybridization here:
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brainly.com/question/1604211</h3><h3 /><h3>#SPJ4</h3>
