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lapo4ka [179]
3 years ago
15

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

8 0
4 years ago
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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:

7 0
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:

a

Explanation:

a is the answer i too

k the test

4 0
3 years ago
Read 2 more answers
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