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2 years ago

14

Anhydrous CoCl2 is blue; CoCl2•6H2O is red. When CoCl2•6H2O is heated and water is clearly being driven off, it becomes uniforml

y violet before turning blue. How can this be explained?

1 answer:

Sedaia [141]2 years ago

4 0

Answer: On losing 6 moles of water, cobalt chloride forms unstable violet-coloured ions, before generating its stable blue-coloured anhydrous form.

Explanation:

The hydrated cobalt chloride loses its 6 water of crystallization, then dissociates into ions: cobalt ions and chlorine ions that appear violet, and quickly combined to form the stable anhydrous Cobalt chloride with blue colour.

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For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions i

kupik [55]

Answer :

Covalent compound : It is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound.

The covalent compound are usually formed when two non-metals react.

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

All the polyatomic ions always form an ionic compound.

Polyatomic ions : It is a charged species that composed of two or more atoms and these charged species are bonded by the covalent bond.

For the given options:

Option A: $KClO_4$

This compound is formed by the combination of potassium, $K^+$ which is a metal and $ClO_4^-$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option B: $Mg(C_2H_3O_2)_2$

This compound is formed by the combination of magnesium, $Mg^{2+}$ which is a metal and $C_2H_3O_2^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option C: $H_2S$

Hydrogen and sulfur, both are non-metals and they will form a covalent compound.

Option D: $Ag_2S$

This compound is formed by the combination of silver, $Ag^{+}$ which is a metal and sulfur, $S^{2-}$ which is a non-metal. Thus, it will form an ionic compound.

Option E: $N_2Cl_4$

Nitrogen and chlorine, both are non-metals and they will form a covalent compound.

Option F: $Co(NO_3)_2$

This compound is formed by the combination of cobalt, $Co^{2+}$ which is a metal and $NO_3^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

6 0

2 years ago

In redox half-reactions, a more positive standard reduction potential means I. the oxidized form has a higher affinity for elect

qwelly [4]

Answer:

The 1st and 4th options are correct

I.the oxidized form has a higher affinity for electrons

IV. the greater the tendency for the oxidized form to accept electrons

Explanation:

Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.

Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.

(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.

4 0

2 years ago

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

2 years ago

93.2 mL of a 2.03 M potassium fluoride (KF) solution

Marrrta [24]

Answer:

1.98 M

Explanation:

Given data

Initial volume (V₁): 93.2 mL

Initial concentration (C₁): 2.03 M

Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

$93.2mL \times \frac{1L}{1000mL} = 0.0932 L$

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

$V_2 = 0.0932L + 3.92 L = 4.01L$

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

$C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M$

3 0

2 years ago

Read 2 more answers

Acetaldehyde shows two UV bands, one with a lmax of 289 nm ( 5 12) and one with a lmax of 182 nm ( 5 10,000). Which is the n p*

UkoKoshka [18]

Answer:

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

Explanation:

The two types of acetaldehyde transition are as follows:

n→π* and π→π*

From the attached diagram we have to:

ΔEn→π* < ΔEπ→π*

ΔEα(1/λ)

Thus:

λn→π* > λπ→π*

In n→π* spin forbidden, the intensity is low. Thus, the molar extinction E for n→π* is very low.

The same way, for π→π* spin allowed the intensity is high. Thus, the molar extinction coefficient E for π→π* is high too.

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

5 0

2 years ago