1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
NNADVOKAT [17]
2 years ago
12

¿por qué la conducción ocurre más lentamente en los gases que en los sólidos?​

Chemistry
1 answer:
SIZIF [17.4K]2 years ago
6 0

Answer:

escribe esa pregunta en una región de Latinoamérica

You might be interested in
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 
8 0
3 years ago
Read 2 more answers
A gas effuses 4.0 times faster than oxygen (o2). what is the molecular mass of the gas? 1.0 g/mol 1.0 g/mol 2.0 g/mol 2.0 g/mol
Anastaziya [24]

Answer:

32(molecular mass has no unit )

Explanation:

(16)(o2)

16×2

=32

5 0
2 years ago
HELPPPPPPPP <br> I NEEED TO PASS
Monica [59]
I believe it might be c I might wrong
7 0
3 years ago
Read 2 more answers
The value of ΔH° for the reaction below is -6535 kJ. ________ kJ of heat are released in the combustion of 16.0 g of C6H6 (l)?
Umnica [9.8K]

Answer:

the value of H° is below -6535 kj. +6H2O

Explanation:

6H2O answer solved

3 0
3 years ago
assuming nitrogen behaves like an ideal gas, what volume would 14.0 g of nitrogen gas (N2) occupy at STP? the gas constant is 0.
dimaraw [331]

Answer:

V = 22.41 L

Explanation:

Given data:

Mass of nitrogen = 14.0 g

Volume of gas at STP = ?

Gas constant = 0.0821 atm.L/mol.K

Solution:

Number of moles of gas:

Number of moles = mass/molar mass

Number of moles= 14 g/ 14 g/mol

Number of moles = 1 mol

Volume of gas:

PV = nRT

1 atm × V = 1 mol × 0.0821 atm.L/mol.K  × 273 K

V = 22.41 atm.L / 1 atm

V = 22.41 L

4 0
3 years ago
Other questions:
  • What part of the space shuttle provides the force needed to take off
    12·1 answer
  • What is the total number of electrons that can occupy the f sub level
    9·2 answers
  • It is important that water is a powerful solvent because ____?
    5·1 answer
  • What is normal body temperature in degrees celsius? express your answer numerically in degrees celsius?
    6·1 answer
  • The density of a sample of gold is 19.32 g/cm'. If the sample has a volume of 48.9 cm', what is the mass?
    15·1 answer
  • How many energy level does carbon have
    6·2 answers
  • When an atom gains an electron to become an ion what happens to its size?
    5·1 answer
  • When using a calorimeter, the initial temperature of a metal is 70.4C. The initial temperature of the water is 23.6C. At the end
    5·1 answer
  • Which notation represents the largest atomic radius?
    12·1 answer
  • A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!