Answer:
The answer to your question is : letter B. 0.25 atm
Explanation:
To solve this problem we need to use the combined gas law:
<u>P₁V₁</u> = <u>P₂V₂</u>
T₁ T₂
Data
P1 = 0.99 atm V1 = 2 l T1 = 273K
P2 = ? V2 = 4 l T2 = 137K
Now, the clear P2 from the equation and we get
P2 = P1V1T2 / T1V2
Substitution P2 = (2 x 0.99 x 137)/(273 x 4)
P2 = 271.26 / 1092
Result P2 = 0.248 atm ≈ 0.25 atm
For this item, we need to assume that air behaves like that of an ideal gas. Ideal gases follow the ideal gas law which can be written as follow,
PV = nRT
where P is the pressure,
V is the volume,
n is the number of mols,
R is the universal gas constant, and
T is temperature
In this item, we are to determine first the number of moles, n. We derive the equation,
n = PV /RT
Substitute the given values,
n = (1 atm)(5 x 10³ L) / (0.0821 L.atm/mol.K)(0 + 273.15)
n = 223.08 mols
From the given molar mass, we calculate for the mass of air.
m = (223.08 mols)(28.98 g/mol) = 6464.9 g
<em>ANSWER: 6464.9 g</em>
Answer:
the stronger light 5.5 m apart from the total illumination
Explanation:
From the problem's statement , the following equation can be deducted:
I= k/r²
where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality
denoting 1 as the stronger light and 2 as the weaker light
I₁= k/r₁²
I₂= k/r₂²
dividing both equations
I₂/I₁ = r₁²/r₂²=(r₁/r₂)²
solving for r₁
r₁ = r₂ * √(I₂/I₁)
since we are on the line between the two light sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus
r₂ = r₁ + d
then
r₁ = (r₁ + d)* √(I₂/I₁)
r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)
r₁*(1-√(I₂/I₁)) = d*√(I₂/I₁)
r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁)) =
r₁ = d/[√(I₁/I₂)-1)]
since the stronger light is 9 times more intense than the weaker
I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3
then since d=11 m
r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m
r₁ = 5.5 m
therefore the stronger light 5.5 m apart from the total illumination
Answer:
The mole fraction of codeine is 5.4%
Explanation:
Mole fraction = Mole of solute / Total moles (Mole solute + Mole solvent)
Solute (Codeine)
Molar mass 299.36 g/m
Mass / Molar mass = Mole → 46.85 g /299.36 g/m = 0.156 moles
Solvent (Ethanol)
Molar mass 46.07 g/m
125.5 g / 46.07 g/m = 2.724 moles
Mole fraction (Codeine) 0.156 / (0.156 + 2.724) → 0.054
Answer:
See below
Explanation:
The half-cell reduction potentials are
Ag⁺(aq) + e⁻ ⇌ Ag(s) E° = 0.7996 V
Co²⁺(aq) + 2e⁻ ⇌ Co(s) E° = -0.18 V
To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.
We get the cell reaction
:
Anode: Co(s) ⇌ Co²⁺(aq) + 2e⁻
<u>Cathode: 2Ag⁺(aq) + 2e⁻ ⇌ 2Ag(s) </u>
Overall: Co(s) + 2Ag⁺ (aq) ⇌ Co²⁺(aq) + 2Ag(s)
