Answer:
All are correct
Explanation:
1) The angular momentum quantum number, l, are the subshells within a shell (principle quantum number) it talks about the "form" of an orbital, the number itself tells you about the number of angular nodes (a plane without electronic density). It starts at l=0 where you don't see any nodes and it takes the form of an sphere, and we knowing it bu another name an s-orbital. It takes values up to n-1.
l=0 (sphere - s-orbital)
l=1 (p-orbital)
l=2 (d-orbital)
2) The magnetic quatum number, ml relates to the number of orbitals within a subshell then it is related with l, taking values form -l to l incluing 0.
For l=0 (s-orbital) ml=0
For l=1 (p-orbital) ml=1,0,-1
For l=2 (d-orbital) ml=2,1,0,-1,-2
3) In every shell we are restricted by the total number of nodes of any orbital. Then if we want a d-orbital with l=3 we need at least 3 plane nodes only achievable with n=3 at least.
Answer is: -601,2 kJ/mol
Chemical reaction: Mg(OH)₂ → MgO + H₂O.
ΔHrxn = 37,5 kJ/mol.
ΔHf(Mg(OH)₂) = <span>−924,5 kJ/mol.
</span>ΔHf(H₂O) = <span>−285,8 kJ/mol.
</span>ΔHrxn -enthalpy of reaction.
ΔHf - enthalpy of formation.
<span>ΔHrxn=∑productsΔHf−∑reactantsΔHf.
</span>ΔHf(MgO) = -924,5 kJ/mol - (-285,8 kJ/mol) + 37,5 kj/mol.
ΔHf(MgO) = -601,2 kJ/mol.
Answer:
please brainlist answer
Explanation:
The addition of K 3 Fe(CN) 6 to a solution causes the formation of a deep blue precipitate which indicates that iron(II) ions are present.
Answer:
THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.
Explanation:
Equation:
Al3+ + 3e- -------> Al
3 F of electricity is required to produce 1 mole of Al
3 F of electricity = 27 g of Al
If 18 g of aluminium was used, the quantity of electricity to be used up will be:
27 g of AL = 3 * 96500 C
18 G of Al = x C
x C = ( 3 * 96500 * 18 / 27)
x C = 193 000 C
For 18 g of Al to be produced, 193000 C of electricity is required.
To calculate the current required to produce 193 000 C quantity of electricity, we use:
Q = I t
Quantity of electricity = Current * time
193 00 = I * 1.50 * 60 * 60 seconds
I = 193 000 / 1.50 * 60 *60
I = 193 000 / 5400
I = 35.74 A
The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A
