What are the two fundamental portions of an atom?

1 answer:

4 0

Number of Protons in an Uncharged Atom
The two main components of an atom are the nucleus and the cloud of electrons. The nucleus contains positively charged and neutral subatomic particles, whereas the cloud of electrons contains tiny negatively charged particles.

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The equation below shows the products formed when a solution of silver nitrate (AgNO3) reacts with a solution of sodium chloride

d1i1m1o1n [39]

I would say that the answer has to be C
Since there is no change in mols on both sides of the equation the mass is constant

3 0

3 years ago

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Why does the Periodic Table have that separated part (lanthanides and actinides) at the bottom?

zepelin [54]

In the periodic table the lanthanoid and the actinides are place separately at the bottom because of their electronic configuration and their properties compared to the other elements.

The the lanthanoid and the actinides are place separately at the bottom in the periodic table due to their electronic configuration and the properties. and to make the periodic table more convenient . if we place the f block elements that is he lanthanoid and the actinides then the size of the periodic table will increase. the f block elements are called as the inner transition element.

Thus , to make the periodic table more convenient and to group the elements in the block the the lanthanoid and the actinides are place separately at the bottom.

To learn more about lanthanoid and actinide here

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3 0

1 year ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$