The carbon-oxygen single bond in a carboxylic acid
Answer:
ΔS = -0.1076 kJ /kg*K
Explanation:
Step 1: Data given
Initial state = 0.8 m³/kg and 25 °C = 298.15 K
Final state = 0. 3³/kg and 287 °C = 560.15 K
Cv = 0.686 kJ/kg*K
Step 2: Calculate the average temperature
The average temperature = (25°C + 287 °C)/2 =156 °C ( = 429 K)
Step 3: Calculate the ΔS
ΔS =(Cv, average) * ln(T2/T1) + R*ln(V2/V1)
ΔS = 0.686 * ln(560.15/298.15) + 0.2598*ln( 0.1/0.8)
ΔS = -0.1076 kJ /kg*K
Answer:
marine biology
coastal ecology
astronomy
meteorology
Explanation:
As a college student, to study oceanography one will have to take classes in the field of marine biology, coastal ecology, astronomy and meteorology.
An oceanographer is someone or a professional that studies the ocean in order have more scientific knowledge about them.
Oceanography is a merger of geology, biology, chemistry, physics as it pertains to the ocean.
- There is no need to study human anatomy to study oceanography.
- Marine biology and coastal ecology deals with study of life forms in their marine environment.
- Astronomy and meteorology helps to gain insight about the formation of the ocean and how weather relates to the ocean.
Answer: The answer is A. A conductor that allows electricity to flow easily
Answer:
150ml
Explanation:
For this question,
NaOH completely dissociates. It is a strong base
HCl also completely dissociates. It is a strong acid
So we have this equation
m1v1 = m2v2 ----> equation 1
M2 = 2m
V1= ??
M2 = 6m
V2 = 50m
When we input these into equation 1, we have:
2m x v1 = 6m x 50ml
V1 = 6m x 50ml/2
V1 = 300/2
V1 = 150ml
Therefore NaOH that is required to neutralize the solution of hydrochloric acid is 150ml.
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