I would say it is the amount of mass, but double check
Ca(OH)2 -------> CaO + H2O
ratio is 1 : 1
CaO Mr= 56g
15/56 = 0.268 mol (mols of CaO in 15g)
Ca(OH)2 Mr= 74g
0.268x74= 19.83g (required amount of Cash(OH)2)
Answer:
1.346 meters per second
Explanation:
You traveled 35 meters in 26 seconds
35 / 26 = avg speed.
35 / 26 = 1.346...
Answer:
Eqv Pt pH = 8.73
Explanation:
HOAc + NaOH => NaOAc + H₂O
50ml(0.10M HOAc) + 50ml(0.10M NaOH) => 100ml(0.05M NaOAc) + H₂O
For neutralized system, 100ml of 0.05M NaOAc remains
NaOAc => Na⁺ + OAc⁻
Na⁺ + H₂O => No Rxn
OAc⁻ + H₂O => HOAc + OH⁻
C(i) 0.05M ----- 0M 0M
ΔC -x ----- +x +x
C(f) 0.05-x
≅ 0.05M ----- x x
Kb = Kw/Ka = [HOAc][OH⁻]/[OAc⁻] = 1 X 10⁻¹⁴/1.7 X 10⁻⁵ = (x)(x)/(0.05M)
=> x = [OH⁻] = SqrRt(0.05 x 10⁻¹⁴/1.7 x 10⁻⁵) = 5.42 x 10⁻⁶M
=> pOH = -log[OH⁻] = -log(5.42 x 10⁻⁶) = 5.27
pH + pOH = 14 => pH = 14 - pOH = 14 - 5.27 = 8.73 Eqv Pt pH
Given question is incomplete. The complete question is as follows.
The successive ionization energies of a certain third-period element are I1 = 577.9KJ/mol, I2 + 1820 KJ/mol, I3 = 2750 KJ/mol, I4 = 11600 KJ/mol, I5 = 14800 KJ/mol. what element do these ionization energies suggest? Explain your reasoning.
Explanation:
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
Here, given ionization energies belong to an element present in third period.
We know that second ionization energy will always be greater than third ionization energy.
For the given ionization energies, there is a huge difference between third and fourth ionization energy. This means that there are three valence electrons present in the element.
This is because after losing three electrons it is difficult for the given element to lose fourth electron. Hence,
is high as compared to
.
Hence, this element has 3 valence electrons and it belongs to 3A group of the periodic table.
Thus, we can conclude that the given unknown element is aluminium (Al).