All categories

3 years ago

Are these right? U guys r the best on this app!!

Chemistry

2 answers:

Pepsi [2]3 years ago

8 0

Yes, they are right.Very good hand writing! You did a very great job on showing your work.

baherus [9]3 years ago

8 0

Yes they are your good a this

You might be interested in

Which of the following is NOT a characteristic of most ionic compounds?

jekas [21]

What are the answer choices???

7 0

3 years ago

Read 2 more answers

Is cl metal nonmetal or metalloid<br> or metalloid

Korvikt [17]

Answer:

nonmetal

Explanation:

go to ptable.com. it helps a lot

4 0

3 years ago

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

3 years ago

Determine the acid dissociation constant for a 0.0250 M weak acid solution that has a pH of 2.37 . The equilibrium equation of i

Oksanka [162]

Answer:

For part (a): pHsol=2.22

Explanation:

I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.

So, you're dealing with formic acid, HCOOH, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.

You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes

HCOOH(aq]+H2O(l]⇌ HCOO−(aq] + H3O+(aq]

I 0.20 0 0

C (−x) (+x) (+x)

E (0.20−x) x x

You need to use the acid's pKa to determine its acid dissociation constant, Ka, which is equal to

3 0

3 years ago

A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached

lara [203]

Answer:

a) cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

b) 0.068 V.

Explanation:

A) Cu2+ + 2e- euilibrium cu (s)

Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-

Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)

B) To calculate the cell voltage

E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+

putting values we get

= 0.339V + (90.05916V/2)log(0.100) = 0.309V

E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.

6 0

3 years ago