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asambeis [7]
3 years ago
10

In a particular reaction between copper metal and silver nitrate, 12.7 g cu produced 38.1 g ag. what is the percent yield of sil

ver in this reaction? cu + 2agno3 → cu(no3)2 + 2ag
Chemistry
2 answers:
mrs_skeptik [129]3 years ago
6 0

88.4% !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

LekaFEV [45]3 years ago
4 0
Answer:  
For these problems, we need to compare the theoretical yield that we'd get from performing stoichiometry to the actual yield stated in the problem. % yield is the actual yield/theoretical yield x 100%  
Cu + 2 AgNO₠→ Cu(NOâ‚)â‚‚ + 2 Ag ==> each mole of copper yields two moles of silver  
12.7-g Cu x ( 1 mol Cu /63.5-g Cu) x ( 2 mol Ag / 1 mol Cu) x (108-g Ag / 1 mol Ag) = 43.2-g Ag. This is the theoretical yield. Now, since we got 38.1-g Ag our % yield is: 
38.1-g/43.2-g x 100% = 88.2%
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What is the energy of a wave with wavelength of 4.2 x 10-7 m. (Hint: Calculate for frequency first.)
erik [133]

Answer:

Option B. 4.74×10¯¹⁹ J.

Explanation:

The following data were obtained from the question:

Wavelength (λ) = 4.2×10¯⁷ m

Energy (E) =.?

Next, we shall determine the frequency of the wave. This can be obtained as follow:

Wavelength (λ) = 4.2×10¯⁷ m

Velocity (v) = constant = 3×10⁸ m/s

Frequency (f) =.?

v = λf

3×10⁸ = 4.2×10¯⁷ × f

Divide both side by 4.2×10¯⁷

f = 3×10⁸ / 4.2×10¯⁷

f = 7.143×10¹⁴ s¯¹

Therefore, the frequency of the wave is 7.143×10¹⁴ s¯¹.

Finally, we shall determine the energy of the wave using the following formula

E = hf

Where

E is the energy.

h is the Planck's constant

f is the frequency

Thus, the enery of the wave can be obtained as follow:

Frequency (f) = 7.143×10¹⁴ s¯¹.

Planck's constant = 6.63×10¯³⁴ Js

Energy (E) =..?

E = hf

E = 6.63×10¯³⁴ × 7.14×10¹⁴

E = 4.74×10¯¹⁹ J

Therefore, the energy of the wave is 4.74×10¯¹⁹ J.

5 0
3 years ago
This is agree or disagree I need help ASAP!!
PIT_PIT [208]
All of them are agree :)
8 0
3 years ago
Read 2 more answers
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
Lucy wants to attach a goal cost to each of her life goals. Why might she do this?
Natasha2012 [34]

Answer:

So she can have something to reach or look forward to.

Explanation:

none

8 0
3 years ago
An atom of titanium has 22 protons and 26 neutrons. If the atoms loses one electron, what will be the charge on the ion that for
Alja [10]
It’s +21 (c) if i answered late my apologies
5 0
3 years ago
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