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4 years ago

Give six examples of complex compounds.

Chemistry

1 answer:

ra1l [238]4 years ago

3 0

Answer:

Examples of complex compound include potassium ferrocyanide K4[Fe(CN)6] and potassium ferricyanide K3[Fe(CN)6]. Other examples include pentaamine chloro cobalt(III) chloride [Co(NH)5Cl]Cl2 and dichlorobis platinum(IV) nitrate [Pt(en)2Cl2](NO3)2.

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Write balanced equations and expressions for ksp for the dissolution of each of the following ionic compounds. Write a balanced

Lina20 [59]

Based on the dissolution equation of BaSO4, the solubility product, Ksp, expression of BaSO4 is: [Ba^{2+}] × [SO4^{2-}]

<h3>What is solubility product Ksp of a solute?</h3>

The solubility product, Ksp, of a substance is an equilibrium constant which expresses the product of the concentration of ions produced by a solute.

The solubility of a solute depends on temperature, thus solubility product also depends on temperature.

The balanced equation for the dissolution of barium sulfate is given below:

BaSO4 ⇌ Ba^{2+} + SO4^{2-}

The solubility product for barium sulfate is given as;

Ksp = [Ba^{2+}] × [SO4^{2-}]

Therefore, the solubility product, Ksp, expression of BaSO4 is: [Ba^{2+}] × [SO4^{2-}]

Learn more about solubility product at: brainly.com/question/26405952

3 0

2 years ago

A certain ionic compound X has a solubility in water of 40.3 g/L at 20. degrees C. Calculate the mass X of required to prepare 5

tino4ka555 [31]

Answer:

20.1 g

Explanation:

The solubility indicates how much of the solute the solvent can dissolve. A solution is saturated when the solvent dissolved the maximum that it can do, so, if more solute is added, it will precipitate. The solubility varies with the temperature. Generally, it increases when the temperature increases.

So, if the solubility is 40.3 g/L, and the volume is 500 mL = 0.5 L, the mass of the solute is:

40.3 g/L = m/V

40.3 g/L = m/0.5L

m = 40.3 g/L * 0.5L

m = 20.1 g

7 0

3 years ago

Please help me with this question, and if u can do it on paper

nasty-shy [4]

1. O2 is the limiting reagent

2. Excess NH3 = 0.102 moles

3. 0.088 mol of NO is formed

4. 2.376 grams of H2O is formed

<h3>Stoichiometric problem</h3>

From the equation: $4NH_3 + 5O_2 --- > 4NO + 6H_2O$

Mole ratio of NH3 and O2 = 4:5

Mole of 3.25 g NH3 = 3.25/17 = 0.19 mol

Mole of 3.50 g O2 = 3.5/32 = 0.11 mol

Equivalent mole of O2 from NH3 = 5/4 x 0.19 = 0.24 mol

Equivalent mole of NH3 from O2 = 4/5 x 0.11 = 0.088 mol

Thus, O2 is the limiting reagent.

Excess NH3 = 0.19 - 0.088 = 0.102 moles

Mole ratio of O2 and NO = 5:4

Equivalent mole of NO = 4/5 x 0.11 = 0.088 mol

Mole ratio of O2 and H2O = 5:6

Equivalent mole of H2O = 6/5 x 0.11 = 0.132 mol

Mass of 0.132 mol H2O = 0.132 x 18 = 2.376 grams

More on stoichiometric problems can be found here: brainly.com/question/15047541

#SPJ1

8 0

2 years ago

Which is larger? 12 milligrams or 12 kilograms

Oxana [17]

Answer: Kilograms

Explanation:

8 0

3 years ago

Read 2 more answers

What mass of carbon dioxide is formed when 1.75 mol of ethane burns completely in oxygen?

dybincka [34]

Answer:

There is a mass of 154 Grams of Carbon Dioxide.

Explanation:

One mole is equal to 6.02 × 10^23 particles.

This means we have 1.05 X 10^24 total particles of Ethane.

Each ethane particle contains 2 carbon atoms.

If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)

Carbon Dioxide has a molar mass of 44.01 g/mol

So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.

Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.

What you should end up with is 154 Grams of Carbon Dioxide.

Hope this helps (And more importantly I hope I didn't make any errors in my math lol)

As a side note this is all assuming that this takes place at STP conditions.

3 0

4 years ago