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almond37 [142]
3 years ago
12

Part C Number of molecules in 8.437x10-2 mol C6H6

Chemistry
1 answer:
N76 [4]3 years ago
5 0

Answer:

There are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

Explanation:

In this problem, we need to find the number of molecules in 8.437\times 10^{-2} mol of C_6H_6.

The molar mass of C_6H_6 is 6\times 12+1\times 6=78\ g/mol

No of moles = mass/molar mass

We can find mass from above formula.

m=n\times M\\\\m=8.437\times 10^{-2}\ mol\times 78\ g/mol\\\\m=6.58\ g

Also,

No of moles = no of molecules/Avogadro number

N=n\times N_A\\\\N=8.437\times 10^{-2}\times 6.023\times 10^{23}\\\\N=5.08\times 10^{22}\ \text{molecules}

Hence, there are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

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marysya [2.9K]

Answer:

Stoichiometry

Explanation:

You have to use the stoichiometry rules. Remember that concentracion is the same as moles divided by the volume in the system

C = n/V

Also, that amount of moles is equal to mass divided by the molarmass

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3 years ago
Classify each of these reactions pb(no3)2+nicl2=pbcl2+ni(no3)2
maxonik [38]

The reaction Pb(NO3)2(aq) + NiCl2(aq)---------> PbCl2(s) + Ni(NO3)2(aq) is a precipitation reaction.

A chemical reaction is said to occur when two or more substances are combined to produce new substances. Chemical reactions are classified based on the kind of change taking place in the reaction.

In the reaction;

Pb(NO3)2(aq) + NiCl2(aq)---------> PbCl2(s) + Ni(NO3)2(aq)

We can see that a solid is formed when lead II nitrate reacts with nickel II chloride. This is a precipitation reaction.

Learn more: brainly.com/question/5624100

4 0
3 years ago
A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

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Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
3 years ago
¿Qué es una fuente de luz primaria?
Tems11 [23]

Answer:

Las fuentes de luz primarias son aquellas que emiten la luz que producen. Por ejemplo: la luz de Sol, un relámpago, la de una vela,...

A su vez las fuentes primarias las podemos clasificar en: fuentes naturales y fuentes artificiales

Explanation:

3 0
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alexgriva [62]
Answer is: <span>the pressure of the gas is 9,2 atm.
</span>p₁ = 4,0 atm.
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4 atm · 5,5 L ÷ 300 K  = p₂ · 2,0 L ÷ 250 K.
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p₂ = 9,2 atm.

4 0
3 years ago
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