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yan [13]
2 years ago
8

Write balanced equation for:

Chemistry
1 answer:
mario62 [17]2 years ago
4 0

Answer:

hmmm good luck

Explanation:

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What is the solution to the problem to the correct number of significant figures (102,900/12)+(170•1.27)
GarryVolchara [31]

As per as the Multiplication rules of the significant figures, whenever any numbers in the decimals forms are multiplied or divided then result in mentioned in such a way so that the significant figures after the decimal will be same as that in the given least condition.


_______________________________


102900/12 = 8575


170 × 1.27 = 215.9


∴ (102,900 ÷ 12) + (170 × 1.27) =  8575 + 215.9


= 8790.9


Now, As per as Above rules, answer in correct significant figures will be = 8791.



8 0
3 years ago
Answer my my science homework please
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What is the concentration of NaCI in an aqueous solution that contains 0.032 grams of NaCI in 600. Grams of the solution
jolli1 [7]

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0.00032 Grams of NaCl per 1 gram of the solution

Explanation:

8 0
3 years ago
Draw the curved arrows to show the mechanism for the reaction of butanedioic (succinic) anhydride with methanol.
Marta_Voda [28]
Here is the mechanism of said reaction.

8 0
3 years ago
A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W
Agata [3.3K]

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

E^0_{[Ag^{+}/Ag]}=+0.80V

E^0_{[Cu^{2+}/Cu]}=+0.34V

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : Cu(s)\rightarrow Cu^{2+}(aq)+2e^-

Reduction half reaction (Cathode) : Ag^{+}(aq)+e^-\rightarrow Ag(s)

Thus the overall reaction will be,

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

7 0
3 years ago
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