The atomic mass would not change since the mass of an electron is negligible compared to the mass of protons and neutrons
Answer:
b) +2 and +3.
Explanation:
Hello,
In this case, given the molecular formulas:
And:
We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:
Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:
Thus, answer is b)+2 and +3
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The compound FeCl3 is made of one atom and three atoms of chlorine
<u><em>explanation</em></u>
- FeCl3 is made up of two elements.
- that is iron with Fe chemical symbol and chlorine with Cl chemical symbol.
- There are 3 atoms of chlorine in FeCl3 because the subscript in front of Cl in FeCl3 is 3.
Answer:
0. 414
Explanation:
Octahedral interstitial lattice sites.
Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.
The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.
The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.
We know that to relate solutions of with the factors of molarity and volume, we can use the equation:
**
NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M
is the left side of the equation. Then we have:
We can then solve for
:
and
or
We now know that the total amount of volume of the 4.35 M solution will be
210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.