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Serjik [45]
3 years ago
15

What is the mass (in grams) of 9.42 × 1024 molecules of methanol (CH3OH)?

Chemistry
2 answers:
Furkat [3]3 years ago
8 0
(9.42x10^24)/(6.02x10^23)=15.6478 mol methanol
15.6478mol(32.04g/1 mol)=501.3568g methanol
Sloan [31]3 years ago
6 0

Answer: 500 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given molecules}}{\text {avogadro's number}}=\frac{9.42\times10^{24}}{6.023\times 10^{23}}=15.6moles

1 mole of CH_3OH weighs = 32.04 g/mol

Thus 15.6 moles of CH_3OH weigh=\frac{32.04}{1}\times 15.6=500g

Thus the mass of 9.42\times10^{24} molecules of methanol is 500 grams

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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) +
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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.

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rewona [7]

Answer:

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