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snow_lady [41]
3 years ago
10

There are 24 g of carbon in a substance with a mass of 30 g. What is the percent carbon by mass for this substance?

Chemistry
1 answer:
mihalych1998 [28]3 years ago
8 0

Answer:

Explanation:

The composition of the vapor in mass percent.

According to Dalton's law,

where  is the partial pressure of the gas j,  is the mole fraction of the gas j in the gas mixture and  is the total pressure.

To know the composition of the vapor we must first calculate the molar fractions of the components of the mixture in the vapor:

→  

 →  

If we consider 1 mole of solution then we will have 0.56 mol of benzene and 0.44 mol of toluene in the vapor. In this way, the number of grams of each component in the vapor will be,

The percentage by mass of each component in the vapor will be,

% benzene = (g benzene / g total) x 100% = (43.7 g / (43.7 g + 40.6 g)) x 100% → % benzene = 51.8 %

% toluene = 100 % - % benzene → % toluene = 41.2%

So, the composition of the vapor in mass percent is 51.8 % benzene and 41.2% toluene.

Why is the composition of the vapor different from the composition of the solution?

The composition of the vapor will be different from that of the solution, since the more volatile compound will have a larger molar fraction in the vapor phase than in the liquid phase.

In a mixture with different volatile components the compound that volatilizes more easily is the one that will have greater capacity to escape from the solution in the form of vapor and, therefore, will be in a greater composition in the vapor above the solution.

We can measure this by means of vapor pressure, which is a measure of the volatility of a substance, that is, the capacity of the substance to pass from a liquid to a gaseous state. In the question, benzene is in greater proportion than toluene in the vapor mixture since it has a higher vapor pressure (94.2 torr) than toluene (28.4 torr).

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Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).
Degger [83]
a) before addition of any KOH : 

when we use the Ka equation & Ka = 4 x 10^-8 : 

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

           = 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

       = 9.2 x 10^-5 M

when PH = -㏒[H+]

   PH = -㏒(9.2 x 10^-5)

        = 4  

b)After addition of 25 mL of KOH: this produces a buffer solution 

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]


first, we have to get moles of HCIO= molarity * volume

                                                           =0.21M * 0.05L

                                                           = 0.0105 moles

then, moles of KOH = molarity * volume 

                                  = 0.21 * 0.025

                                  =0.00525 moles 

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L =  0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

                                        = 0.00525 / 0.075

                                        =0.07 M

and molarity of KCIO = moles KCIO / total volume

                                    = 0.00525 / 0.075

                                    = 0.07 M

and when Ka = 4 x 10^-8 

∴Pka =-㏒Ka

         = -㏒(4 x 10^-8)

         = 7.4 

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4 

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume 

                                                        = 0.21 M * 0.05L

                                                        = 0.0105 moles

then moles KOH = molarity * volume
                            =  0.22 M* 0.035 L 

                            =0.0077 moles 

∴ moles of HCIO remaining = 0.0105 - 0.0077=  8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume 

                                      = 8 x 10^-5 / 0.085

                                      = 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

                                          = 0.0077M / 0.085L

                                          = 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

                                   = 0.0105mol / 0.1 L

                                   = 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

       = 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO] 

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

           = 3.8
∴PH = 14- POH

        =14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

 when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M 

V2 is the excess volume added  of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

           = -㏒0.02 

           = 1.7

∴PH = 14- POH

       = 14- 1.7 

      = 12.3 
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3 years ago
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What are the complete Ionic and Net Ionic equations for the following
zhenek [66]

Answer:

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Explanation:

5 0
3 years ago
Why is it better to measure the percent change in mass<br> ratherthan just using the change in mass?
never [62]

Answer: The result is presented in proportion which gives a clearer understanding and accurate result.

Explanation: Percentage change in mass is the proportion of the initial mass of a substance changed after sometime. The results is presented as a percentage making it more accurate and can help to give future reference to weight calculations.

Change is Mass is the mass of a substance left after sometime mostly given in grams. It is not as accurate as percentage change in mass. It is generally better to show results in percentage change in mass as it gives a better understanding of what mass of a substance was lost after a given period or after application of energy like Heat or increased temperature.

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At exactly 2:00 PM, speedy the hermit crab crawl onto a meter stick starting at the 10 cm mark. If he reaches the 60 cm mark at
kvv77 [185]

Answer:

speed=5cm/minute

Explanation:

Given at 2:00PM the hermit is at 10cm mark

Also given that the hermit reaches 60cm mark at 2:10PM

Therefore the time elapsed is  10 minutes = 10\times 60=600 seconds

the distance travelled by the hermit is 60-10=50cm=0.5m

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