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olganol [36]
3 years ago
10

The first IE of neon (atomic number 10) is significantly higher than that of argon (atomic number 18) but significantly

Chemistry
1 answer:
lbvjy [14]3 years ago
8 0

Explanation:

Different atoms binds their outermost shell electrons with different amount of energy.

The amount of energy required to remove an electron from an atom is the ionization energy.

  • Ionization energy measures the readiness of an atom to lose electrons.
  • From the given problem, we can infer that in group O the ionization energy decreases down the group.
  • Helium has the highest ionization energy.
  • Down a group on the periodic table, ionization energy decrease because:
  1.  atomic radii increases down the group.
  2. there is an increasing shielding/screening effect of inner shell electrons on the outermost shell electrons.

Learn more:

Ionization energy brainly.com/question/2153804

#learnwithBrainly

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Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for
eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0
4 years ago
In a study of the decomposition of the compound XX via the reaction
Leya [2.2K]

Answer:

( About ) 0.03232 M

Explanation:

Based on the units for this reaction it should be a second order reaction, and hence you would apply the integrated rate law equation "1 / [X] = kt + 1 / [X_o]"

This formula would be true for the following information -

{ X_o = the initial concentration of X, k = rate constant, [ X ] = the concentration after a certain time ( which is what you need to determine ), and t = time in minutes }

________

Therefore, all we have left to do is plug in the known values. The initial concentration of X is 0.467 at a time of 0 minutes, as you can tell from the given data. This is not relevant to the time needed in the formula, as we need to calculate the concentration of X after 18 minutes ( time = 18 minutes ). And of course k, the rate constant = 1.6

1 / [X] = ( 1.6 )( 18 minutes ) + 1 / ( 0.467 ) - Now let's solve for X

1 / [X] = 28.8 + 1 / ( 0.467 ),

1 / [X] = 28.8 + 2.1413...,

1 / [X] = 31,

[X] = 1 / 31 = ( About ) 0.03232 M

Now for this last bit here you probably are wondering why 1 / 31 is not 0.03232, rather 0.032258... Well, I did approximate one of the numbers along the way ( 2.1413... ) and took the precise value into account on my own and solved a bit more accurately. So that is your solution! The concentration of X after 18 minutes is about 0.03232 M

3 0
4 years ago
How many atoms are in 20.4 g of beryllium
Xelga [282]
<h3>Answer: sorry i don't know what the answer that question </h3>

8 0
3 years ago
What is the purpose of fractional distillation of dicyclopentadiene in this experiment?
eimsori [14]
Fractional distillation is a process by which components in a chemical mixture are separated into different parts (called fractions) according to their different boiling points. Fractional distillation is used to purify chemicals and to separate mixtures to obtain their components.
3 0
4 years ago
Only electrons in the __________ orbital can form chemical bonds.
slavikrds [6]

Answer:

Electrons in sigma bonds remain localized between two atoms. Sigma bond results from the formation of a molecular orbital by the end to end overlap of atomic orbitals. Electrons in pi bonds can become delocalized between more than two atoms. Pi bonds result from the formation of molecular orbital by side to side overlap of atomic orbitals.

6 0
3 years ago
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