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iogann1982 [59]
3 years ago
14

How many milliliters of 0.132 m hclo4 solution are needed to neutralize 50.00 ml of 0.0789 m naoh?

Chemistry
1 answer:
Murljashka [212]3 years ago
3 0

Answer: 29.89mL of HClO4 is needed

Explanation: Please see the attachments below

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Help please please please please
Rom4ik [11]

Answer:

D. 18.7 grams

Explanation:

The coefficients are the key.

Create a proportion with them, and the molar mass, and then solve for x:

\frac{2}{3}  =  \frac{x}{28}  \\ 3x = 2(28) \\ 3x = 56 \\ x = 18.666... = 18.7

7 0
3 years ago
How does the number of coupons you use affect the amount of money you save?
slega [8]
Using coupons seems like an old-fashioned, time-consuming way to save a few cents on the products you buy at the grocery store. After all, most coupons don't usually have a face value of more than $1 and, quite often, they're worth far less. But done properly, couponing can save you thousands of dollars every year.
7 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
Calculate the oxidation state of the underlined elements of K2 Cr2 O7
Semenov [28]

Answer:

+4

Explanation:

In PbO2, oxygen exhibits an oxidation number of -2 (since it's not a peroxide or superoxide):

Let the oxidation number of Pb be x. Then, for the compound to be neutral, the oxidation numbers of all atoms should add up to zero.

⇒ x + (−2) + (−2) = 0

x = +4

So the oxidation no. of Pb is +4.

I hope this helps.

3 0
2 years ago
Whose main job is to help a plant cell make proteins
hoa [83]

Answer:

Water

Explanation:

Because water is very important

7 0
3 years ago
Read 2 more answers
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