Answer:
kindly check the EXPLANATION SECTION
Explanation:
In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio of any nuclei ranges from to 1 to 1.5, then we say the nuclei is STABLE.
Also, we need to understand that when the Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.
So, let us check which is stable and which is unstable:
a. 4 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/4= stable.
b. 7 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/7= 1 = stable.
c. 2 protons and 3 neutrons = Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.
d. 3 protons and 0 neutrons = Neutron- proton ratio = N/P = 0/3= 0= unstable.
e. 6 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.
f. 9 protons and 9 neutrons = Neutron- proton ratio = N/P = 9/9 = 1 = stable.
g. 8 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/8 =0.875 = unstable.
h. 1 proton and 0 neutrons = Neutron- proton ratio = N/P = 0/1 =0 = unstable
B, homeostasis. “Homeostasis is the state of steady internal, physical, and chemical conditions maintained by living systems.”
Simply put, MA = Force Out / Force in. That's the way it is usually stated. The force out is normally what you need to move. The force in is what you need to supply to get the force out. Most machines will give you an MA of more than 1. Some (like your arm) will give you less than 1 and others (like this one) will give you exactly one.
This one is frictionless, otherwise it would slip into less than one if it had friction.
Answer B
The structure will be:
H₃C-CH₂-CH=CH-CH₂-CH₃
This class of compounds is known or referred to as alkenes. Alkenes are unsaturated hydrocarbons that contain a carbon-carbon double bond. The present of this double bond alters the properties of alkenes rom alkanes.