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Verizon [17]
3 years ago
7

The following ions contain the same number of electrons. Rank them in order of decreasing ionic radii. Rank from largest to smal

lest radius. To rank items as equivalent, overlap them. se3+ cl-
Chemistry
1 answer:
solong [7]3 years ago
8 0

Cl- < Se3+ is the decreasing order of ionic radii of the two ions.

Explanation:

Atomic number of Se= 34 Se3+ will have 31 electrons (+ sign indicates loss of electron)

Atomic number of chlorine is 17, Cl- will have 18 electrons (- sign shows gain of electrons)

Number of electrons in their valence shell:

Se3+ = 2,8,18,3

Cl- = 2,8,8

The Se3+ atoms have K.L.M.N shells while Cl- K,L,M shells

The size of the ions is based on the effective nuclear charge on the ion. The loss in electron causes small cation and gain in electron causes increase in anion size.

The number of occupied shells more the number more is the ionic radii.

Se3+ will have large ionic radii as Se+ has four filled shells.

Cl- < Se3+ (decreasing order of ionic radii)

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The half-life of radon-222 is 3.8 days. If a sample currently contains 3.1 grams of radon-222, how much radon-222 did this sampl
vodka [1.7K]

The   number  of grams of  radon 222 did it have 15.2  ago was  49.6 grams( answer  C)

 <u>calculation</u>

  • calculate the number of half life it has covered    from   15.2  days to 3.8 days

             that is  divide 15.2/ 3.8 = 4 half life

  •  half  life is time taken for a radio activity  of a specified isotope to fall to half its original mass

therefore  3.8 days ago  it was  3.1 x2 =  6.2 grams

                  7.6 days ago  it was 6.2 x2 = 12.4 grams

                   11.4  days  ago it was 12.4  x2=  24.8  grams

                  15.2  days  ago  it was  24.8 x2=49.6 grams





5 0
3 years ago
What substances can be separated into simpler substances only by chemical means?
Mnenie [13.5K]

A compound is a substance that can be separated into simpler substances only by chemical means.

6 0
3 years ago
1. Nitrogen, oxygen, and fluorine atoms all seem pretty similar, at first. How many valence electrons does an atom of each have?
matrenka [14]
Nitrogen is 5 valence electrons
oxygen is 6 valence electrons
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6 0
3 years ago
When excess dilute hydrochloric acid was added to sodium sulphite 960 of sulphuric (iv) oxide was produced. calculate the mass o
irakobra [83]

The mass of sodium sulfite that was used will be 1,890 grams.

<h3>Stoichiometric problems</h3>

First, the equation of the reaction:

NaSO_3 + 2HCl --- > NaCl_2 + H_2O + SO_2

The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.

Mole of 960 grams SO2 = 960/64 = 15 moles

Equivalent mole of sodium sulfite that reacted = 15 moles

Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

3 0
1 year ago
A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C
laila [671]

Answer:

  • First choice: 99 J/K

Explanation:

<u>1) First law of thermodynamic (energy balance)</u>

  • Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter

<u>2) Energy change of each substance:</u>

  • General formula:

     

Heat released or absorbed = mass × Specific heat × change in temperature

  • density of water: you may take 0.997 g/ ml as an average density for the water.

  • mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g

  • Specif heat of water: 1 cal / g°C

  • Heat released by the hot water:

       Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)

  • Heat absorbed by the cold water:

       Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)

  • Heat absorbed by the calorimeter

       Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)

<u>4) Balance</u>

  • Heat₁ = Heat₂ + Heat₃

49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)

  • Solve for Ccal

Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K

Ccal = 23.6 cal/ K

  • Convert to cal / K to Joule / K

  • 1 cal = 4.18 Joule

       23.6 cal / K × 4.18 J / cal = 98.6 J/K

Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.

4 0
3 years ago
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