Answer:
1. 67.2 kJ/mol
Explanation:
Using the derived expression from Arrhenius Equation
Given that:
time 
= 8.3 days = (8.3 × 24 ) hours = 199.2 hours
time 
= 10.6 hours
Temperature 
= 0° C = (0+273 )K = 273 K
Temperature 
= 30° C = (30+ 273) = 303 K
Rate = 8.314 J / mol
Since 
Then we can rewrite the above expression as:
Answer:
a. 5.0 x 10⁷ brown grains = 50 million
b. 5.0 x 10³ brown grains = 5000
Explanation:
The concentration of 2 % brown sand means we have for every 100 grains of sand 2 are brown.
We need to calculate the number of brown sand in the bucket as follows:
= 2.5 x 10⁹ billion grains of sand x (2 brown grains/ 100 grains of sand)
= 5.0 x 10⁷ brown grains
Likewise if the concentration of brown sand is 2.0 ppm, it mean that we have 2 brown grain per every million grains of sand.
= 2.5 x 10⁹ billion grains of sand x ( 2.0 brown grains/10⁶ grains of sand )
= 5.0 x 10³ brown grains
The answers make sense since a concentration of 1 part per million is ten thousandths of a 1 percent
Answer:
2.9 is the initial pH of the analyte solution.
Explanation:
The dissociation constant of acetic acid as per theoretical value = 
The initial concentration of acetic acid = c = 0.0900 M
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
![K_a=\frac{[Ac^-][H^+]}{[HAc]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BAc%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHAc%5D%7D)
Solving for x:
x = 0.001264 M
![[H^+]=0.001264 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.001264%20M)
The pH of the solution :
![pH=-\log[0.001264]=2.898\approx 2.9](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B0.001264%5D%3D2.898%5Capprox%202.9)
2.9 is the initial pH of the analyte solution.
Answer:
22.77 g.
he limiting reactant is O₂, and the excess reactant is Mg.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (16.3 g) of Mg and (4.52 g) of oxygen:
no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.
no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.
So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.
<em>∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).</em>
<u><em>Using cross multiplication:</em></u>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.565 mol of Mg produce → <em>0.565 mol of MgO.</em>
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.565 mol)(40.3 g/mol) = <em>22.77 g.</em>
