Fine particles, ground level ozone, sulfur dioxide, nitrogen dioxide, lead
Answer:
V₂ = 0.95 L
Explanation:
Given data:
Initial temperature of gas = 171.4 K
Final temperature of gas = 288.4 K
Final volume = 1.6 L
Initial volume = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₁ = V₂T₁ /T₂
V₂ = 1.6 L × 171.4 K / 288.4 k
V₂ = 274.24 L.K / 288.4 K
V₂ = 0.95 L
Molar mass C6H12O6 = 12 x 6 + 1 x 12 + 16 x 6 = 180 g/mol
1 mole ------------ 180 g
( moles ) --------- 843.211 g
moles = 843.211 x 1 / 180
moles = 843.211 / 180
= 4,684 moles of C6H12O6
Answer:
0.86 g/mL
Explanation:
In order to solve the problem, you have to know the formula for "Density." Density refers to<em> </em><em>"mass per unit volume." </em><em>It's formula is:</em>
- <em>Density (ρ) = </em>
<em />
Next, we have to convert the units of measurement in accordance to what is being asked in the problem. It is asking for "g/mL," which means we have to convert 1.25 quarts<em> (qt.) </em>to mL and 2.25 pounds<em> (lbs.)</em> to grams.
Let's solve.
1.25 qt. x 
=<em> 1,182.94 mL</em>
2.25 pounds x 
= <em>1,020.58 grams</em>
Now that we converted them to their appropriate units of measurement, we can now solve for density.
- <em>ρ = </em><em />
- <em>ρ = </em>
<em />
<em>The density of the motor oil is 0.86 g/mL</em>
t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half life of the element and λ is decay constant.
32 = 0.693 / λ
λ = 0.693 / 32 (1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, λ is decay constant and t is the time taken.
t = 1.9 hours = 1.9 x 60 min
From (1) and (2),
Nt = Nο e⁻Λ(0.693/32)*1.9*60
Nt = 0.085Nο
Percentage = (Nt/Nο) x 100%
= (0.085Nο/Nο) x 100%
= 8.5%
Hence, Percentage of remaining atoms with the original sample is 8.5%
