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1 year ago

What is formula for chromium (III) phosphate trihydrate and cobalt (II) phosphate octahydrate

2 answers:

7 0

The formula for chromium (III) phosphate trihydrate is CrPO4- 3H20. This compound if in the anhydrous state, exists as a green crystal whereas a hydrated form violet crystal. The formula for cobalt(II) phosphate octahydrate is Co3(PO4)2•8H2O.

lozanna [386]1 year ago

6 0

Answer:

1. Chromium (III) phosphate trihydrate: CrPO₄•3H₂O

2. Cobalt (II) phosphate octahydrate: Co₃(PO₄)₂•8H₂O

Explanation:

Chromium (III) phosphate is an inorganic chemical compound having the chemical formula CrPO₄, in which the chromium metal is present in the +3 oxidation state. The anhydrous chromium (III) phosphate is green in color.

The trihydrate chromium (III) phosphate has the chemical formula CrPO₄•3H₂O.

Cobalt (II) phosphate, also called cobalt violet, is an inorganic chemical compound having the chemical formula Co₃(PO₄)₂, in which cobalt is present in +2 oxidation state.

The octahydrate cobalt (II) phosphate has the chemical formula Co₃(PO₄)₂•8H₂O

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The ksp of zinc carbonate (znco3 is 1.0 × 10–10. what is the solubility concentration of carbonate ions in a saturated solution

Trava [24]

Ksp - solubility product constant is equivalent to equilibrium constant, except this constant is used to determine the solubility of ions of a solid in a solution.
ksp is the product of the soluble ions in the compound. Higher the ksp value, higher the degree of solubility.
ZnCO₃ (s) ---> Zn²⁺ (aq) + CO₃²⁻ (aq)
n n
ksp = [Zn²⁺][CO₃²⁻]
In the equation equal amounts of ions Zn²⁺ and CO₃²⁻ ions are soluble.
amount of ions soluble = n
ksp is therefore equal to;
ksp = n x n
ksp = n²
ksp = 1 * 10⁻¹⁰ M
therefore
1 * 10⁻¹⁰ M = n²
n = 1 x 10⁻⁵ M
therefore concentration of CO₃²⁻ = 1 x 10⁻⁵ M

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2 years ago

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How many milliliters of 0.02 M HCl are needed to react completely with 100 mL of 0.01 M NaOH?

serious [3.7K]

The reaction of HCl and NaOH is HCl + NaOH = NaCl + H2O. So the mole number of HCl and NaOH is equal. So the volume of HCl =0.01*0.1/0.02=0.05 L =50 ml. So the answer is D).

3 0

2 years ago

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An aluminum ion has a charge of +3, and an oxide has a charge of -2. What would be the product of a reaction between these two e

kozerog [31]

The product of a reaction between these two elements is $Al_{2} O_{3}$ .

Explanation:

The oxidation state of an ion in a compound is equal to its charge.

The aluminum having a charge of +3 because oxidation state is +3

The oxide is having charge of -2

The product of these reactants will produce a chemical compound.

The compound formed is $Al_{2} O_{3}$ i.e Aluminium oxide. The compound while getting formed will share the charge and cation A+ will have the charge of anion and anion will have the charge of cation. This will result in a compound as there should be a neutral charge on the compound formed.

The +3 charge of the cation Al+ will go to anion oxide O2- and the charge of anion -2 will go with cation Al+. 

8 0

2 years ago

In a post office, a 3-m long ramp is used to move carts onto a dock that is higher than 1 m. Which describes how the IMA of this

pishuonlain [190]

Answer:

well I think the answer is it depends on the friction

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2 years ago

Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f

Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0

2 years ago