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ivolga24 [154]
3 years ago
11

How many electrons are transferred between the cation and anion to form the ionic bond in one formula unit of each compound? (1

electrons, 2 electrons, 3 electrons, 4 electrons) ?NaCICaSBaOKBrLiF
Chemistry
1 answer:
kodGreya [7K]3 years ago
4 0

The number of transferred electrons to form the ionic bond in one formula unit of each compound is as below

NaCl = 1 electron

CaS = 2 electrons

BaO= 2 electrons

KBr=1 electrons

LIF = 1 electron



explanation

ionic bond is formed between a metal and a nonmetal . This bond is formed when electron are transferred between cation and anion. To know how many electrons transferred you look the oxidation state of the metal.


for example 

in NaCl only one electrons is transferred since Na is in oxidation state of +1. Na donate the one electron to Cl to form an ionic bond.  Ba  is in oxidation state  of 2+ and hence it  donate the two electrons to O to form ionic bond.

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mol/L

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0.71 J/g°C

Explanation:

Here is the complete question

thermometer A 51.9 g sample of quartz is put into a calorimeter (see sketch at right) that contains 300.0 g of water. The quartz sample starts off at 97.8 °C and the temperature of the water starts off at 17.0 °C. When the temperature of the water stops changing it's 19.3 °C. The pressure remains constant at 1 atm. insulated container water sample Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to 2 significant digits. a calorimeter g °C

Solution

Since the temperature of the water increases from 17.0 °C to 19.3 °C, it means that it loses heat. Also, the final temperature of the quartz equals the final temperature of the water 19.3 °C. Since the quartz temperature decreases from 97.8 °C to 19.3 °C it loses heat.

So, heat lost by quartz, Q = heat gained by water, Q'

-Q = Q'

-mc(θ₂ - θ₁) = m'c'(θ₂ - θ₃) where m = mass of quartz = 51.9 g, c = specific heat capacity of quartz, θ₁ = initial temperature of quartz = 97.8 °C, θ₂ = final temperature of quartz = 19.3 °C, m' = mass of water = 300 g, c = specific heat capacity of water = 4.2 J/g °C , θ₃ = initial temperature of water = 17.0 °C, θ₂ = final temperature of water = 19.3 °C

Making c subject of the formula, we have

c = -m'c'(θ₂ - θ₃)/m(θ₂ - θ₁)

Substituting the values of the variables into the equation, we have

c = -300 g × 4.2 J/g °C(19.3 °C - 17.0 °C)/51.9 g(19.3 °C - 97.8 °C)

c = -1260 J/°C(2.3 °C)/51.9 g(-78.5 °C)

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