Correction: The temperature change is from 20 °C to 30 °C.
Answer:
Cp = 1.0032 J.g⁻¹.°C⁻¹
Solution:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 5016 J
m = mass = 500 g
Cp = Specific Heat Capacity = ??
ΔT = Change in Temperature = 30 °C - 20 °C = 10 °C
Solving eq. 1 for Cp,
Cp = Q / m ΔT
Putting values,
Cp = 5016 J / (500 g × 10 °C)
Cp = 1.0032 J.g⁻¹.°C⁻¹
Answer:
Following are the responses to the given choices:
Explanation:
- The RBC crenation is implied through NaCl by 2,67 percent(m/v) because that solution becomes hypertonic to RBC because of the water within the RBC that passes externally towards the outskirts. RBC thus shrinks.
- 1.13% (m/v), because the low concentration or osmotic that all this solution shows is hypotonic regarding RBC because of the water which has reached the resulting swelling in RBC.
- Distilled H2 implies hemolytic distillation.
- Glucose is indicated by crenation at 8.69 percent (m/v).
- 5.0% (m/v) glucose and 0.9% (m/v) (Crenation is indicated by NaCl.v)
The answer is (3) Cu2O. Copper (I) has an oxidation state of +1 (that's what the "I" indicates). You can also think of this as copper (I) having a charge of +1. Oxygen has an oxidation state of -2 (that's just a rule you have to know), and you can think of it as oxygen having a charge of -2. You need oxidation numbers in a neutral compound to add up to 0 (or charges in a neutral compond to add up to 0), so you need two Cu to balance the O, which is Cu2O.
C. Fixing a wineglass requires a large decrease in entropy.
