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3 years ago

Just help please And does anybody know Johnny from NCT

Chemistry

2 answers:

Reil [10]3 years ago

6 0

Yes the name of it is

kvv77 [185]3 years ago

4 0

YES NCTZEN!!!

HELLO MYSELF MRS NA JAEMIN (≧▽≦)

#STR.E.AM. NCTDREAM "HELLO FUTURE"

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I need help balance equations BaCl2+NaOH=NaCl+Ba(OH) 2

gtnhenbr [62]

Answer:

BaCl2+ 2NaOH=2NaCl+Ba(OH)2

Explanation:

If you need an explanation on how I balanced it then let me know

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4 years ago

sergeinik [125]

Rigid crust perhaps

7 0

3 years ago

Any two uses of mixtures

romanna [79]

Answer:

Any two used of mixtures are 1) sand and water 2) sugar and salt

Explanation:

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3 years ago

A concentration cell is built based on the following half reactions by using two pieces of zinc as electrodes, two Zn2+ solution

Lana71 [14]

Explanation:

The given reaction at cathode will be as follows.

At cathode: $Zn^{2+} + 2e^{-} \rightarrow Zn$ , $E_{o}$ = -0.761 V

At anode: $Zn \rightarrow Zn^{2+} + 2e^{-}$ , $E_{o}$ = 0.761

Therefore, net reaction equation will be as follows.

$Zn^{2+} + Zn \rightarrow Zn + Zn^{2+}$

Initial: 0.129 - - 0.427

Change: -0.047 - - -0.047

Equilibrium: (0.129 - 0.047) (0.427 - 0.047)

= 0.082 = 0.38

As $E^{o}_{cell}$ for the given reaction is zero.

Hence, equation for calculating new cell potential will be as follows.

E_{cell} = $E^{o}_{cell} - \frac{RT}{nF} ln \frac{[Zn^{2+}]_{products}}{[Zn^{2+}]_{reactants}}$

= $0 - \frac{8.314 atm L/mol K \times 291 K}{2 \times 96500} ln \frac{0.38}{0.082}$

= 0.019

Thus, we can conclude that the cell potential of the given cell is 0.019.

3 0

4 years ago

Help me!

AnnZ [28]

Answer:

Explanation:

a is the answer i too

k the test

4 0

3 years ago

Read 2 more answers