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Irina18 [472]
1 year ago
10

What is the IUPAC name for this product?

Chemistry
1 answer:
mihalych1998 [28]1 year ago
7 0

The IUPAC name for the given product is 2 chloro Butane.

<h3>What is IUPAC nomenclature?</h3>

IUPAC stands for 'International Union of Pure and Applied Chemistry', which givers some rule for designing the name of compounds of chemistry.

  • In the given product total four carbon atoms are present and between all of them single bonds are present.
  • In the second carbon atom, chlorine group is present.
  • During the nomenclature process, first we write down the name of the attached group which is followed by the alkane chain.

Hence name of the product is 2 chloro Butane.

To know more about IUPAC nomenclature, visit the below link:

brainly.com/question/26635784

#SP1

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Answer:

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Explain why you hear a “whoosh” sound when you open a can containing a carbonated drink. Which gas law applies?
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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
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I need help with number 8 please help ASAP.
Rzqust [24]

Answer:

33.33% = 33%

Explanation:

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We need to get the number of mole of CO2:

and when we have 0.22 g of CO2, so number of mole = mass / molar mass

Moles = 0.22 g / 44 g/mol = 0.005 mole

Moles of Mg = moles of CO2 = 0.005 mole

Mass of Mg = moles * molar mass

= 0.005 * 84 /mol = 0.42 g

Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100

=33.33 %

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