# of atoms per mol = Avogadro’s # (6.022 x 10^23)
Number of mols = mass of substance / molar mass
73 g / 40.08 g = 1.8 mols of Ca in 73 grams
1.8 mols x avagadro’s # = 1.1 x 10^24 atoms in 73 grams of Ca
Answer:
Increase
Explanation:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
If the initial temperature and pressure is standard,
Pressure = 1 atm
Temperature = 273.15 K
then we increase the temperature to 400.0 K, The pressure will be,
1 atm / 273.15 K = P₂/400.0K
P₂ = 1 atm × 400.0 K / 273.15 K
P₂ = 400.0 atm. K /273.15 K
P₂ = 1.46 atm
Pressure is also increase from 1 atm to 1.46 atm.
Answer:
uhhh i do not know this but we are learning at my school i could give you the answer soon i you want
Explanation:
Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.
Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.
Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.
Let the molar solubility be S upon dissociation.
PbBr₂ or Lead Bromide dissociates in pure water as follows:
PbBr₂ ----------> Pb⁺² + Br⁻
S 2S
Ksp = [Pb⁺²] [ Br⁻]
Ksp = (S) (2S)²
Ksp = 4S³
6.60 × 10⁻⁶ = 4S³
S = 0.0118M
Hence, the Molar solubility S is 0.0118M.
Learn more about Molar solubility here, brainly.com/question/16243859
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Explanation:
Moles of metal,
=
4.86
⋅
g
24.305
⋅
g
⋅
m
o
l
−
1
=
0.200
m
o
l
.
Moles of
H
C
l
=
100
⋅
c
m
−
3
×
2.00
⋅
m
o
l
⋅
d
m
−
3
=
0.200
m
o
l
Clearly, the acid is in deficiency ; i.e. it is the limiting reagent, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal.
So if
0.200
m
o
l
acid react, then (by the stoichiometry), 1/2 this quantity, i.e.
0.100
m
o
l
of dihydrogen will evolve.
So,
0.100
m
o
l
dihydrogen are evolved; this has a mass of
0.100
⋅
m
o
l
×
2.00
⋅
g
⋅
m
o
l
−
1
=
?
?
g
.
If 1 mol dihydrogen gas occupies
24.5
d
m
3
at room temperature and pressure, what will be the VOLUME of gas evolved?
