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DiKsa [7]
3 years ago
7

Lines on a map joining places that have the same temperature?

Chemistry
2 answers:
vazorg [7]3 years ago
8 0

Answer:

Isotherms

Explanation:

almond37 [142]3 years ago
4 0

Answer:

it is Isotherms

Explanation:

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The density of osmium, which is the densest metal, is 22.57 g/cm^3. What is the mass of a block of osmium that measures 1.00 cm
olganol [36]
Density = Mass / Volume
V = 1.00 * 4.00 * 2.50 = 10 cm³
22.57 g/cm³ = Mass / 10 cm³
M = 22.57 g/cm³ * 10 cm³
M = 225.7 g
Answer: The mass of the block of osmium is 225.7 g.
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Which of the following phrases best describes a synthesis reaction?
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What is the ph of the benzoic acid solution prior to adding sodium benzoate?
Reika [66]
The acid dissociation constant of benzoic acid is 6.5 x 10^-5. Therefore, the pH of the benzoic acid solution prior to adding sodium benzoate is:

pH = -log[Ka]
pH = -log (6.5 x 10^-5)
pH = 4.19

The pH of the benzoic acid solution is 4.19 which is acidic, but a weak acid. 

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Marina86 [1]

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2 years ago
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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.

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3 years ago
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