Helpppp I need helpp plesse

Object A has a temperature of 287 Kelvin, object B has a temperature of 374 Kelvin. In what direction will thermal energy flow?

Mashcka [7]

Thermal energy will flow from an object high temperature to an object of low one. In this case, the thermal energy will flow from object B to object A.

6 0

3 years ago

Read 2 more answers

Which compound is most likely to follow second-order kinetics in a substitution reaction? (ch3)2chbr ch3ch2br (ch3)3cch2br ch3br

Inga [223]

Substitution reaction<span> monitors second-order kinetics; that is, the degree of the reaction hinge on on the concentration of two first-order reactants. In the circumstance of bimolecular nucleophilic </span>substitution, these two reactants are the nucleophile and the haloalkane. So for this problem, the answer is ch3br.

4 0

4 years ago

What is the name of Ce element in periodic table?

Elden [556K]

Answer:

Cerium

Explanation:

6 0

4 years ago

Read 2 more answers

What is the standard electrode potential for a galvanic cell constructed in the appropriate way from these two half-cells?

____ [38]

E

θ

Cell

=

+

2.115

l

V

Cathode

Mg

2

+

/

Mg

Anode

Ni

2

+

/

Ni

Explanation:

Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]

Mg

2

+

(

a

q

)

+

2

l

e

−

→

Mg

(

s

)

−

E

θ

=

−

2.372

l

V

Ni

2

+

(

a

q

)

+

2

l

e

−

→

Ni

(

s

)

−

E

θ

=

−

0.257

l

V

The standard reduction potential

E

θ

resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (

298

l

K

,

1.00

l

kPa

) is defined as

0

l

V

for reference. [2]

A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.

Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher

E

θ

and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower

E

θ

will experience oxidation and act the anode.

E

θ

(

Ni

2

+

/

Ni

)

>

E

θ

(

Mg

2

+

/

Mg

)

Therefore in this galvanic cell, the

Ni

2

+

/

Ni

half-cell will experience reduction and act as the cathode and the

Mg

2

+

/

Mg

the anode.

The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:

E

θ

cell

=

E

θ

(

Cathode

)

−

E

θ

(

Anode

)

E

θ

cell

=

−

0.257

−

(

−

2.372

)

E

θ

cell

=

+

2.115

Indicating that connecting the two cells will generate a potential difference of

+

2.115

l

V

across the two cells.

5 0

2 years ago

Read 2 more answers

In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O

yarga [219]

HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:

HClO + H2O ⇄ H3O+ +OCl-

Then the equilibrium constant, Ka, of dilute HClO would be:

$K_{a} = \frac{[ H_{3} O^{+} ][O Cl^{-} ]}{HClO}$

Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.

HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x

Substituting the excess (E) concentration to the Ka equation:

$K_{a} = \frac{[x ][x]}{1.0 \ x \ 10^{-4} - x }$

Simplifying the equation would yield a quadratic equation:

$x^{2} + K_{a}x-(1.0 \ x \ 10^{-4}) K_{a}=0$

The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.

x = 1.688 x 10^-6 M = [H3O+] = [ClO-]

Then, you can determine the conc of [OH-] through pH.

pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M

Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M

Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.

Thus in relative amounts, the order would be

H2O >>> HClO > H3O+ = ClO- > OH-

6 0

3 years ago

Read 2 more answers