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Debora [2.8K]
4 years ago
9

Scientists can make atoms of large elements which have not been previously known. Identify an element that would have similar pr

operties to an atom that has 118 protons.
Chemistry
1 answer:
olasank [31]4 years ago
3 0

Answer:

Radon

Explanation:

One element that will have similar property as an atom with 118 protons is Radon because it will belong to the p-block and the noble gas group.

The element will follow atomic number 117 which is already in the 7th group on the periodic table.

  • Generally, on the periodic table, elements in the same group will have the same chemical property.
  • The valency of most elements determines their chemical behavior.
  • Since our mystery element is in the 8th group, noble gas group, it will mostly behave like any of the elements in the group.
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Its a covalent bond for this q
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3 years ago
A learner was assigning oxidation numbers for different elements in the compounds OF2 and NaF. The learner assigned F an oxidati
77julia77 [94]

Answer: B. No, fluorine is always assigned an oxidation number of -1.

Explanation:

6 0
2 years ago
What is the volume of a 5.30g piece of aluminum? The density for aluminum is 2.70mL
Makovka662 [10]

Answer:

1.96mL

Explanation:

Density = mass/volume, and rearranged to solve for volume, volume = mass/density.

So:

volume = 5.30g/2.70g/mL = 1.96mL (assuming your unit was g/mL for density)

7 0
3 years ago
as the elements period 3 are considered in order of increasing atomic number, the number of principal energy levels in each succ
Diano4ka-milaya [45]

Answer:

stay the same.

Explanation:  Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):

Na:  1s2 2s2 2p6 3s1

Ar:    s2 2s2 2p6 3s2 3p6

As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:

Na: 3s1

Ar: 3s2, 3p6

All the new electrons electrons filled the third energy level (3s and 3p).  So the energy level does not change, just the orbitals.

5 0
3 years ago
Write the fraction of the mass of kcl produced from 1 g of k2c03​
Minchanka [31]

Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂

mol of K₂CO₃(MW=138 g/mol) :

= 1 g : 138 g/mol

= 0.00725

From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :

= 2/1 x mol K₂CO₃

= 2/1 x 0.00725

= 0.0145

Mass of KCl(MW=74.5 g/mol) :

= mol x MW

= 0.0145 x 74.5

= 1.08 g

6 0
3 years ago
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