Answer:
Mass = 57.05 g
Explanation:
Given data:
Volume of SO₂ = 20.0 L
Temperature = standard = 273 K
Pressure = standard = 1 atm
Mass of SO₂ = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
n = PV/RT
n = 1 atm × 20.0 L / 0.0821 atm.L/ mol.K× 273 k
n = 20.0 / 22.41/mol
n = 0.89 mol
Mass of SO₂:
Mass = number of moles × molar mass
Mass = 0.89 mol × 64.1 g/mol
Mass = 57.05 g
Answer:
4 mL
Explanation:
Use the density formula, d = m/v
Plug in the density and mass to solve for v:
4 = 16/v
4v = 16
v = 4
So, the volume is 4 mL
1st you pour the mixture<span> through a filter to remove the larger </span>pebbles<span>. Next, add water to dissolve the salt and then filter out the </span>pepper the last thing you do <span>evaporate the water to leave the salt behind.</span>
Answer:
The concentration of cyclobutane after 875 seconds is approximately 0.000961 M
Explanation:
The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M
The final concentration of cyclobutane, [
] = 0.00119 M
The time for the reaction, t = 455 seconds
Therefore, the Rate Law for the first order reaction is presented as follows;
![\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }](https://tex.z-dn.net/?f=%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%20%3D%20%5Ctext%20%7B-k%7D%20%5Ccdot%20t%20%7D)
Therefore, we get;
![k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}} {-t }](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7B%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%7D%20%20%7B-t%20%7D)
Which gives;
k ≈ 5.088 × 10⁻⁴ s⁻¹
The concentration after 875 seconds is given as follows;
[] = [A₀]·
Therefore;
[] = 0.00150 × 
= 0.000961
The concentration of cyclobutane after 875 seconds, [] ≈ 0.000961 M
