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2 years ago

Lab worksheets oh what a tangled web we weave

1 answer:

8 0

Proper format and submission of lab worksheets is helps develop professional standards. Please follow all format and submission guidelines carefully. The guidelines are meant to provide all employees with a standard organizational format for information from in progress and completed lab work. Also providing easy work flow should you have stop any where in process of a lab.

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List the ions in solution in the electrolysis of conc.sodium chloride

Licemer1 [7]

Chloride ions Cl –(aq) (from the dissolved sodium chloride) are discharged at the positive electrode as chlorine gas, Cl 2(g) sodium ions Na +(aq) (from the dissolved sodium chloride) and hydroxide ions OH –(aq) (from the water) stay behind - they form sodium hydroxide solution, NaOH(aq)

5 0

2 years ago

Calculate the concentrations of all species present in a 0.26 M solution of ethylammonium chloride (C2H5NH3Cl).

Alina [70]

Answer:

0.00000223

Explanation:

pKa for C2H5NH3+ = 10.7

pKw = 14.0

pKa + pKb = pKw

10.7 + pKb = 14.0

pKb = 14.0 - 10.7

pKb = 3.30

C2H5NH3Cl is a salt of ethylamine and HCl so it will dissolve in water to produce C2H5NH3^+ + Cl^-

The base hydrolysis reaction: C2H5NH3^+(aq) + H2O(l) <=> C2H5NH2(aq) + H3O^+(aq)

This reaction is described by Kb.

Kb = [C2H5NH2][H3O^+]/[C2H5NH3^+]

Let [C2H5NH2] = [H3O^+] = x,

so [C2H5NH3^+] = 0.26 - x

Kb = x^2/(0.26 - x) = 2.00 x 10^-11

Let's solve for x. In this equation, It is possible to solve without the use quadratic equation. So we can assume that 0.26 - x is approximately equal to 0.26. We won't know until we do the calculation.

We get: x^2 + 2.00 x 10^-11x - 4.99 x 10^-12 = 0

With the use of a quadratic calculator.

x = 2.23 x 10^-6 M = [C2H5NH2] = [H3O^+]

0.26 - x is just 0.26 M in this problem because 2.23 x 10^-6 M is insignificant.

[C2H5NH3^+] = 0.26 M = [Cl^-]

NOTE:

pH = -log [H3O^+] = -log(2.23 x 10^-6) = 5.65

Ka is the acid dissociation constant

Kb is the base dissociation constant

5 0

3 years ago

What is the net charge of the ionic compound calcium fluoride

yKpoI14uk [10]

Answer:

The net charge of the ionic compound calcium fluoride is zero (0).

Explanation:

Ionic compounds, such as covalent ones, have zero net charge; this is, they are neutral.

Substances with net positive charge are cations and substances with net negative charge are anions.

The charges in the ionic compound calcium flouride are distributed in this way:

Compound formula: CaF₂

Calcium charge: Ca²⁺: this is, each calcium ion has a 2 positive charge

Fluoride charge: F⁻: each fluoride ion has a 1 negative charge.

Then, the net charge is: 1 × (2+) + 2 × (1-) = +2 - 2 = 0.

So, a two positve charge, from one calcium ion, is equal to two negative charges, from two fluoride tions, yielding a zero net charge.

8 0

3 years ago

I just need help this is hard

blondinia [14]

Answer:

you can do anything if you put your mind to it. Also, school is not everything in life there are amazinng things in life that you need school for. you got this.

7 0

2 years ago

Melts in the system pb-sn exhibit regular solution behavior. at 473°c apb = 0.055 in a liquid solution of xpb = 0.1. calculate t

Masteriza [31]

Given the temperature 746 K and activity of Pb equal to 0.055. The mole fraction of Pb is 0.1. So, the mole fraction of Sn = 0.9.Activity coefficient, γ = 0.055 / 0.1 = 0.55.The expression for w=ln⁡〖γ_Pb x RT〗/(X_Sn^2 )=(-0.5978 x 8.314 J/(mol K ) x 746 K)/(0.9 x 0.9)= -4577.7 J= -4578 J

Now we use the computed value above and new temperature 773 K. The mole fraction of Sn and Pb are 0.5 and 0.5 respectively. Calculate the activity coefficient in the following manner.lnγ_Sn=w/RT X_Pb^2=(-4578 J)/(8.314 J/mol x 773 K) x 0.5 x 0.5= -0.718lnγ_Sn=exp⁡(-0.178)=0.386The activity of Sn= γ_Sn x X_Sn=0.386 x 0.5=0.418
w of the system is -4578 J and the activity of Sn in the liquid solution of xsn at 500 degree Celsius is 0.418

6 0

3 years ago