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Step2247 [10]
3 years ago
6

19. When 0.5g of calcium trioxocarbonate (IV) was added to excess dilute

Chemistry
1 answer:
Flauer [41]3 years ago
7 0

Answer:

0.1gmin-1

Explanation:

mass = 0.5g

time = 5minutes

rate= mass/time = 0.1gmin-1

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Atoms of metallic elements tend to...
lara31 [8.8K]

The answer is: lose electrons and form positive ions.

Most metals have strong metallic bond, because of strong electrostatic attractive force between valence electrons (metals usually have low ionization energy and lose electrons easy) and positively charged metal ions.

The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).

For example, magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.  

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.  

3 0
3 years ago
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP
Dvinal [7]
Vol.250 before its to much pressure

7 0
3 years ago
Read 2 more answers
What is the mass of 7.675 x 1023 atoms C? Answer in units of g.
nalin [4]

Answer: 15.3 grams C

Explanation:  1 mole is 6.02x10^23 atoms.  We can find the moles of C in 7.675 x 10^23 atoms of C by dividing:

(7.675 x 10^23 atoms C)/(6.02x10^23 atoms C/mole) = 1.275 moles C

The molar mass of carbon is 12g/mole.  So the mass of 7.675 x 10^23 atoms is (1.275 moles C)*(12 g/mole C) = 15.3 grams.

3 0
3 years ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
3 years ago
Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxid
rodikova [14]

Answer:

sodium hexachloroplatinate(IV)- Na2[PtCl6]

dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br

pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2

Explanation:

The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.

The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.

4 0
3 years ago
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