Answer:
0.94
Explanation:
Let's assume that the total sample of the soil is 1 cm³; which has a porosity of 0.4 cm³; then the original sample of the soil will be 0.6 cm³
The mass of the soil particles in 1 cm³ of soil + water = 0.75 g
The fraction of the total atrazine can be calculated by the expression;
Total atrazine =
Also; 
NOTE THAT: 
Total atrazine can now be calculated as:
= 
= 0.94
Thus,the fraction of total atrazine that will be adsorbed to the soil = 0.94.
Answer:
D. Water and Carbondioxide
Explanation:
Combustion of organic compounds in the presence of excess Oxygen will liberate carbondioxide (CO2) and water vapour (H2O). This is an exothermic reaction because heat is liberated to the surroundings.
CnH2n+1OH(aq) + (3/2*n)O2(g) --> nCO2(g) + (n + 1)H2O(g)
Addition of Oxygen can also be termed as a redox reaction. In this case, alcohols are oxidised while the Oxygen is reduced.
Example, (propanol)
C3H7OH(aq) + 9/2O2(g) --> 3CO2(g) + 4H2O(g)
Answer:
<h2>Volume = 9.29 mL</h2>
Explanation:
Density of a substance can be found by using the formula
From the question
Density = 11.3 g/mL
mass = 105 g
Substitute the values into the above formula and solve for the volume
That's
Cross multiply
11.3v = 105
Divide both sides by 11.3
v = 9.29203
We have the final answer as
<h3>Volume = 9.29 mL</h3>
Hope this helps you
Answer:
H2O is the base dissociating for H3O+.
Answer:
6.25 g
Explanation:
From the question given above, the following data were obtained:
Half-life (t½) = 68.8 years
Time (t) = 344 years
Original amount (N₀) = 200 g
Amount remaining (N) =?
Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 68.8 years
Time (t) = 344 years
Number of half-lives (n) =
n = t / t½
n = 344 / 68.8
n = 5
Thus, 5 half-lives has elapsed.
Finally, we shall determine the amount of the Uranium-232 that remains. This can be obtained as follow:
Original amount (N₀) = 200 g
Number of half-lives (n) = 5
Amount remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2⁵ × 200
N = 1/32 × 200
N = 200 / 32
N = 6.25 g
Thus, the amount of Uranium-232 that remains is 6.25 g
