Answer:
D: It will increase because smaller particles provide more surface area to react.
Explanation:
When the large iron is broken up into smaller pieces, there are more places for the iron to react (meaning there's more surface area). Think of it like taking the surface area of a big cube compared to the surface area of a bunch of small cubes. The sum of the surface areas of the small cubes will be greater than that of the large cube. As a result, more places for the iron to react will cause for a greater reaction.
Answer:
Eukarya- Multicellar and unicellar
Explanation:
Fugi- Multicellar
Protista- unicellar
eubacteria- unicellar and have no nucleus
Eukarya is the only option, and that is because Eukarya is both! It contains a nucleus, and a membrane making it the only one that fits into the incomplete graphic! Hope this helps! (I noticed you were waiting for a while so I came to help! <3)
Answer:
0.05257 L/s
Explanation:
Step 1: Given data
The school uses 1200 gallons/day
Step 2: Convert "gal/day" to "L/day"
We will use the conversion factor 1 gal = 3.785 L.
1200 gal/day × (3.785L/gal) = 4542 L/day
Step 3: Convert "L/day" to "L/s"
We will use the following conversion factors:
4542 L/day × (1 day/24 h) × (1 h/3600 s) = 0.05257 L/s
Part A
75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF
This combination will form a buffer.
Explanation
Here, weak acid HF and its conjugate base F- is available in the solution
Part B
150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl
This combination cannot form a buffer.
Explanation
Here, moles of HF = 0.15 x 0.1 = 0.015 moles
Moles of HCl = 0.135 x 0.175 = 0.023
Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution
Part C
165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH
This combination will form a buffer.
Explanation
Moles of HF = 0.165 x 0.1 = 0.0165 moles
Moles of KOH = 0.135 x 0.05 = 0.00675 moles
Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer
Part D
125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl
This combination will form a buffer
Explanation
Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer
Part E
105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
This combination will form a buffer
Explanation
Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles
Moles of HCl = 0.095 x 0.1 = 0.0095 moles
Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
