in order to determine empirical formula we have to determine the mole ratio of the given elements
Let the total mass of the compound is 100g
as given that the compound has 40% sulfur , so mass of sulfur = 40g
as given that the compound has 60% oxygen, so mass of oxygen = 60g
let us calculate the moles of each element
Moles of sulfur = mass / atomic mass = 40 / 32 = 1.25
moles of oxygen = mass / atomic mass = 60/ 16 = 3.75
In order to get simple ratio of moles we will divide both the moles with least number of moles which is 1.25
moles of sulfur = 1.25 / 1.25 = 1
moles of oxygen = 3.75 /1.25 = 3
So empirical formula will be SO₃
Answer:
Since KOH is a strong base, the solution completely ionizes into K+ and OH- when in water. The reaction KOH --> K+ + OH- takes place. The concentration of [ OH- ] can then be used to calculate the pOH of the solution. pH = 14 - pOH = 14 - 1.48 = 12.52
Explanation:
Answer:
option D is correct
Explanation:
no of moles in 3 grams of HCL=3/36=0.08
if 1 mole of HCL require 1 mole of NaOH then 0.08 moles required 0.08 moles of NaOH
mass of 0.08 moles of NaOH=moles*molar mass=0.08*40=3.2 grams
so 3 grams are required in the reaction
Answer:
Rb = +1 , Sr = +2, In= +3, Sn = +4, Sb= +5
Explanation:
Formula:
Zeff = Z - S
Z = atomic number
S = number of core shell or inner shell electrons
For Sn:
Electronic configuration:
Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²
Zeff = Z - S
Zeff = 50 - 46
Zeff = +4
For Rb:
Electronic configuration:
Rb₃₇ = [Kr] 5s¹
Zeff = Z - S
Zeff = 37 - 36
Zeff = +1
For Sb:
Electronic configuration:
Sb₅₁ = [Kr] 4d¹⁰ 5s² 5p³
Zeff = Z - S
Zeff = 51 - 46
Zeff = +5
For In:
Electronic configuration:
In₄₉ = [Kr] 4d¹⁰ 5s² 5p¹
Zeff = Z - S
Zeff = 49 - 46
Zeff = +3
For Sr:
Electronic configuration:
Sr₃₈= [Kr] 5s²
Zeff = Z - S
Zeff = 38 - 36
Zeff = +2
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is
H = mcΔt
H - heat energy
m - mass of material
c - specific heat of the material
Δt - change in temperature
substituting the values in the equation
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
