Answer: 17.6psi
Explanation:
V1 = 10.5 L
P1 = 14.3 psi
V2 = 8.55L
P2 =?
P1V1 = P2V2
14.3 x 10.5 = P2 x 8.55
P2 = (14.3 x 10.5) / 8.55
P2 = 17.6psi
b. Na2HPO4 + NaH2PO4.
A <em>buffer </em>is a solution of a weak acid and its conjugate base. The weak acid is H2PO4^(-) and its conjugate base is HPO4^(2-).
All the other options are incorrect because they consist of only a single component.
Answer:
d. 60.8 L
Explanation:
Step 1: Given data
- Heat absorbed (Q): 53.1 J
- External pressure (P): 0.677 atm
- Final volume (V2): 63.2 L
- Change in the internal energy (ΔU): -108.3 J
Step 2: Calculate the work (W) done by the system
We will use the following expression.
ΔU = Q + W
W = ΔU - Q
W = -108.3 J - 53.1 J = -161.4 J
Step 3: Convert W to atm.L
We will use the conversion factor 1 atm.L = 101.325 J.
-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L
Step 4: Calculate the initial volume
First, we will use the following expression.
W = - P × ΔV
ΔV = - W / P
ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L
The initial volume is:
V2 = V1 + ΔV
V1 = V2 - ΔV
V1 = 63.2 L - 2.35 L = 60.8 L
Answer:
C. If farmers do not use soil conservation practices, the topsoil, which contains humus, will not be fertile to promote plant growth because of the loss of nutrients from the soil.
Explanation:
The top soil is rich in humus and not weathered materials. It is not close to the bedrock. Rather, the topsoil is rich in organic matter and it is present as humus in the soil.
- If crop rotation is not being practiced, the soil's nutrients would be highly depleted without replenishing them back.
- With crop rotation the top soil can be made more fertile when it is left uncultivated for a period of time.
- Instead of heavily relying on a single piece of land, crops can rotated on a timed basis so that soils will have enough time to restore their nutrients.