Answer:
43.868 J
Explanation:
Kinetic energy of a body is the amount of energy possessed by a moving body. The SI unit of kinetic energy is the joule (kg⋅m²⋅s⁻²).
According to classical mechanics, kinetic energy = 1/2 m·v²
Where, m= mass in kg and v= velocity in m/s
Given: m = 19.2 lb and v = 7.10 miles/h
Since, 1 lb= 0.453592 kg
∴ m = 19.2 lb = 19.2 × 0.453592 kg = 8.709 kg
Also, 1 mi = 1609.34 m and 1 h = 3600 sec
∴ v = 7.10 mi/h = 7.10 × 1609.34 m ÷ 3600 sec = 3.174 m/sec
Therefore, <u>kinetic energy of the goose</u> = 1/2 m·v² = 1/2 × (8.709 kg)× (3.174 m/sec)² = 43.868 J
The correct answer would be the fourth option. A nucleotide consists of a phosphate group, a pentose sugar, and a nitrogen containing base that are all linked together by covalent bonds. Nucleotides are the monomer units of nucleic acids and is the basic unit of the DNA.
Answer:
C₆H₁₂O₆ and O₂ are reactant.
CO₂ and H₂O are products.
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + ATP
Explanation:
There are two types of respiration:
1. Aerobic respiration
2. Anaerobic respiration
Aerobic respiration
It is the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.
Glucose + oxygen → carbon dioxide + water + 38ATP
Anaerobic Respiration
It is the breakdown of glucose molecule in the absence of oxygen and produce small amount of energy. Alcohol or lactic acid and carbon dioxide are also produced as byproducts.
Glucose→ lactic acid/alcohol + 2ATP + carbon dioxide
This process use respiratory electron transport chain as electron acceptor instead of oxygen. It is mostly occur in prokaryotes. Its main advantage is that it produce energy (ATP) very quickly as compared to aerobic respiration.
Steps involve in anaerobic respiration are:
Glycolysis
Glycolysis is the first step of both aerobic and anaerobic respiration. It involve the breakdown of one glucose molecule into pyruvate and 2ATP.
Fermentation
The second step of anaerobic respiration is fermentation. It involve the fermentation of pyruvate into lactic acid or alcohol depending upon the organism in which it is taking place. There is no ATP produced, however carbon dioxide is released in this step.
The mass will stay the same because of the conservation of mass
The question is incomplete, complete question is :
Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (
for HF is
.)
[HF] = 0.280 M
Express your answer to two decimal places.
Answer:
The pH of an 0.280 M HF solution is 1.87.
Explanation:3
Initial concentration if HF = c = 0.280 M
Dissociation constant of the HF = 

Initially
c 0 0
At equilibrium :
(c-x) x x
The expression of disassociation constant is given as:
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)


Solving for x, we get:
x = 0.01346 M
So, the concentration of hydrogen ion at equilibrium is :
![[H^+]=x=0.01346 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.01346%20M)
The pH of the solution is ;
![pH=-\log[H^+]=-\log[0.01346 M]=1.87](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.01346%20M%5D%3D1.87)
The pH of an 0.280 M HF solution is 1.87.