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Inessa05 [86]
3 years ago
8

I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even thou

gh solubility rules would not lead one to predict so because potassium nitrate is obviously soluble and so should copper (II) iodide. One can deduce from the formation of a precipitate that copper is reduced. Write a proposed reaction for the oxidation reduction of copper (II) iodide. Justify the choice of the substance that reduces the copper based on experimental evidence. Also, justify the choice using the atomic structure of potassium ion and iodide ion.
Chemistry
1 answer:
Alexus [3.1K]3 years ago
3 0

Answer:

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Explanation:

The reaction performed in the experiment is;

2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2

The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.

The oxidation - reduction equation is as follows;

2Cu2^+ + 2e ----> 2Cu^+ reduction half equation

2I^- ----> I2 + 2e. Oxidation half equation

Balanced redox reaction equation;

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

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When an alkaline earth metal, a, reacts with a halogen, x, the formula of the covalent compund formed should be a2x?
Ierofanga [76]

Alkaline earth metal are the elements present in II group in the periodic table and are known as 'Metals' and have a charge of +2.

Alkaline earth metals - Be , Mg Ca, Sr , Ba, Ra

Halogens are present in VII A group in the periodic table and are 'Non-metals' and have a charge of -1.

Halogens - F, Cl, Br, I, At

When Alkaline earth metal (metals) combine with Halogens (non-metals) the compound formed will be ionic compound and the formula of the compound will be based on the charges of the element.

When we write the formula of the ionic compound the charges of the elements get criss crossed.

For example - Mg (Alkaline earth metal) have a charge of +2 and Cl (Halogen) have a charge of -1 and when they combine to form the formula their charges get criss crossed and we will get Mg_{1}Cl_{2} or MgCl_{2}

When an alkaline earth metal, A, reacts with a halogen, X, the formula of the Ionic compound formed should be AX_{2}


3 0
3 years ago
What would be oil classification?
boyakko [2]
The service rating of passengers car and commercial automotive motor oils
8 0
3 years ago
The rate constant for this second‑order reaction is
o-na [289]

Answer:

daddadaddddddddddddddddddddddddddddda

Explanation:

dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd

3 0
3 years ago
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
What do you put in the calculator to get the molar mass of Mg3P2
olga55 [171]

Input the atomic masses of Mg and P to give 134.84g/mol

Explanation:

The molar mass of a substance (atom or molecule or compound) is the mass in grams of one mole of the substance:

When dealing with an element the molar mass is the relative atomic mass expressed as g/mol.

For compounds, you add the atomic masses of the component atoms and you sum up.

You simply input the atomic mass of 3 atoms of Mg and 2 atoms of P

Atomic mass of Mg = 24.3g/mol

                             P = 30.97g/mole

Molar mass of Mg₃P₂ = 3(24.3) + 2(30.97) = 134.84g/mol

learn more:

Molar mass brainly.com/question/2861244

#learnwithbrainly

6 0
3 years ago
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