The inner core is solid, fourth layer and made out of iron and nickel. It's the one mostly in charge for the other layers. If the inner core stopped spinning, the outer core would lose its magnetic field, and this will be bad because of the sun's radiation wave. Including the other layers.
The outer core is the liquid, third layer. It's in charge of Earth's magnetic field.
The mantle is the second layer of earth, the original temperature can come up about to 1000+ or more, celsius.
M{(NH₄)₂Cr₂O₇}=2A(N)+8A(H)+2A(Cr)+7A(O)
M{(NH₄)₂Cr₂O₇}=252.065 g/mol
M(Cr)=51.996 g/mol
m{(NH₄)₂Cr₂O₇}/M{(NH₄)₂Cr₂O₇}=m(Cr)/2M(Cr)
m(Cr)=2M(Cr)m{(NH₄)₂Cr₂O₇}/M{(NH₄)₂Cr₂O₇}
m(Cr)=2*51.996*35.8/252.065=14.770 g
m(Cr)=14.770 g
Answer:
C: Mg
Explanation:
Hybridization of atomic orbitals is a fundamental concept introduced by Pauling that describes the mixing of orbitals at an atom which adds a definite direction to the Lewis - shared electron pair or electron chemical - bond concept.
Carbon(C) can hybridized on sp, sp2 and sp3 simply because it's valence shell gives room for it.
For silicon(si), when forming covalent bonds with other atoms, it's 3s and 3p orbitals are mixed with each other to form new hybrid orbitals.
Magnesium in itself doesn't hybridized except in magnesium hydrides.
Boron orbitals(B): when boron forms bonds with three other atoms like borazine, they are hybridized to either the sp2 or hybridized to the sp3 which occurs when boron forms bonds with four atoms just as is in metal borohydrides.
. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:
E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ
Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ
2. The wavelength of a photon with this energy would be:
Energy = hc/wavelength
wavelength = hc/energy
wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m
Converting to nanometers gives: 91.16 nm
3. Repeat the calculation in 1, but using n=5.
4. Repeat the calculation in 2 using the energy calculated in 3.
Answer:
2950 izuzjdjdi1vzhkdzhhzizuzhz
