The question is incomplete, the complete question is;
Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75
Answer:
A nucleus with a A nucleus with a neutron:proton ratio of 1.49
A nucleus with a mass of 187 and an atomic number of 75
Explanation:
The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1
Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.
Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.
Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.
Answer:
The disadvantages of each of the given model of electron configuration have been mentioned below:
1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.
2). Arrow and line diagrams make the counting of electrons and take up too much space.
3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.
Answer: The are bodies of rock or gas that are named for ancient gods.
Explanation:
Earth is the THIRD planet in the solar system. It support life of all living organisms.
VENUS is the second planet in our solar system. It is named after the Roman goddess of love and beauty.
MARS is the fourth planet in our solar system. It is named after the Roman goddess of war.
Thus, Venus and Mars, are EARTH NEIGHBORS
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:
It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:
Thus, the mass in excess is:
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