solution:
The quoted atomic mass on the Periodic Table is the WEIGHTED average of the individual isotopic masses. The higher the isotopic percentage, the MORE that isotope will contribute to the isotopic mass. For this reason, most masses that are quoted on the Table are non-integral.
By way of example we could look to the hydrogen atom. The VAST majority of hydrogen atoms (in this universe) are the protium isotope. i.e. 1H, whose nuclei contain JUST the defining proton. There is a smaller percentage (>1%) of hydrogen atoms WITH one NEUTRON in their nuclei to give the deuterium isotope. i.e. 2H, and because this is relatively cheap, and easily incorporated into a molecule, deuterium labelling is routinely used in analysis.
And there is even a smaller percentage of hydrogen atoms with TWO NEUTRONS in their nuclei, to give the tritium isotope. i.e. 3H. The weighted average of the isotopic percentages gives 
The answer is d. The answer is that because both of them can reproduce that way.
Answer:
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).
Explanation:
The reaction between an acid and a base is called neutralization, forming a salt and water.
Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.
When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:
V acid *M acid = V base *M base
where V represents the volume of solution and M the molar concentration of said solution.
In this case:
- V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
- M acid= 0.129 M
- V base= ?
- M base= 0.135 M
Replacing:
0.0137 L* 0.129 M= V base* 0.135 M
Solving:

V base=0.0131 L = 13.1 mL
<u><em>
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>
Answer:
E²⁺
Explanation:
The group two contain alkaline earth metals.
There are six elements in group 2A.
Beryllium, Magnesium, calcium, strontium, barium and radium.
All members have two valance electrons.
They lose two valance electrons to complete the octet.
When they lose the two valance electrons they form cation X²⁺.
They react with halogens and form salt such as
MgCl₂, CaCl₂ etc.
Mg²⁺ Cl²⁻₂
The oxidation state of halogens are -1, while the elements of group two A shows +2 that's why two atoms of halogen are combine with one atom of alkaline earth metals and make the compound overall neutral.
All the alkaline earth metals have similar properties.
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