<span>A cation is an atom that loses a valence electron. When a valence electron is released there is one electron less to create a repulsive force. The loss of a repulsive force will allow the atom to pull tighter together. An anion would therefore be larger in size due to increased repulsion of the valence electrons.</span>
Answer:
3.91 moles of Neon
Explanation:
According to Avogadro's Law, same volume of any gas at standard temperature (273.15 K or O °C) and pressure (1 atm) will occupy same volume. And one mole of any Ideal gas occupies 22.4 dm³ (1 dm³ = 1 L).
Data Given:
n = moles = <u>???</u>
V = Volume = 87.6 L
Solution:
As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....
= ( 1 mole × 87.6 L) ÷ 22.4 L
= 3.91 moles
<h3>2nd Method:</h3>
Assuming that the gas is acting ideally, hence, applying ideal gas equation.
P V = n R T ∴ R = 0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹
Solving for n,
n = P V / R T
Putting values,
n = (1 atm × 87.6 L)/(0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹ × 273.15K)
n = 3.91 moles
Result:
87.6 L of Neon gas will contain 3.91 moles at standard temperature and pressure.
Scientists use scientific notation to simplify numbers, basically. When dealing with really big numbers or really small numbers, the usage of scientific notation prevents them from having to write a bunch of zeroes.
Hope that helped you!
Answer:
D
Explanation:
Metallic character decreases as you move across a period in the periodic table from left to right. This occurs as atoms more readily accept electrons to fill a valence shell than lose them to remove the unfilled shell. Metallic character increases as you move down an element group in the periodic table. This is because electrons become easier to lose as the atomic radius increases, where there is less attraction between the nucleus and the valence electrons because of the increased distance between them.
We are given the chemical reaction and the amount of reactant used for the process. We use these data together to obtain what is asked. We do as as follows:
0.882 mol H2O2 ( 1 mol O2 / 2 mol H2O2 ) = 0.441 mol O2 produced
Hope this answers the question.
