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kenny6666 [7]
3 years ago
15

Will mark branliest. Calculating percent yield: For the reaction Cu + 2AgNO3 → Cu(NO3)2 + 2 Ag, adding 127 grams of copper to lo

ts of silver nitrate yields 410 grams of silver. What is the percent yield of silver?
I'm desperate.
Chemistry
1 answer:
pishuonlain [190]3 years ago
8 0

Percent yield = (Actual Yield/ Theoretical Yield) x 100

Given, Actual Yield of silver = 410 grams

Given, mass of copper = 127 g

Atomic mass of copper = 63.546 amu

Formula: Moles = mass / atomic mass

Moles of copper = 127 g/ 63.546 amu = 1.998 mol

Based on the given balanced chemical reaction, the molar ratio between Cu: Ag is 1:2

So 1.998 mol of copper should yield (2 mol Ag/ 1 mol Cu) x 1.998 mol of Cu = 3.996 mol

Calculated mol of Ag = 3.996 mol

Atomic mass of silver = 107.8682 amu

Mass of silver = moles x atomic mass = 3.996 mol x 107.8682 amu = 431 g

Based on the math, the theoretical yield = 431 g

Percent yield of silver = (410g/431g) x 100 = 95. 13%

The answer is 95. 13%

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You weigh out 0.1183 g of a complex salt to analyze for the percentage of cyanide ion in your complex salt. After dissolving the
Stels [109]

Answer:

25.35%

Explanation:

Again let me restate the the equation of the reaction;

H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)

Amount of potassium permanganate  reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles

If 2 moles of MnO4 - reacts with 3 moles of CN-

8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2

= 1.229 * 10^-3 moles of CN-

Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol

= 0.03 g

Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100

= 25.35%

8 0
2 years ago
hydrolysis of decapeptide P with the enzyme trypsin affords the following fragments: Glu-Gly-Lys, Gln-Val-Ile, Ala-Ser-Phe-Lys.
ehidna [41]

Answer:

The sequence of an amino acid P is:

Glu-Gly-Lys-Ala-Ser-Phe-Lys-Gln-Val-Ile

Explanation:

Fragments obtained on hydrolysis of decapeptide P by the action of an enzyme named trypsin:

  • Glu-Gly-Lys,
  • Gln-Val-Ile
  • Ala-Ser-Phe-Lys

Fragments obtained on hydrolysis of decapeptide P by the action of an enzyme named chymotrypsin:

  • Lys-Gln-Val-Ile,
  • Glu-Gly-Lys-Ala-Ser-Phe

In order to determine the sequence of protein P , we will arrange fragments in such a way so that common fragments or the common parts of fragments should come under each other.

On arranging these fragments :

Glu-Gly-Lys-Ala-Ser-Phe

Glu-Gly-Lys

                   Ala-Ser-Phe-Lys

                                         Lys-Gln-Val-Ile

                                                Gln-Val-Ile

The sequence of an amino acid P is:

Glu-Gly-Lys-Ala-Ser-Phe-Lys-Gln-Val-Ile

3 0
3 years ago
I need help! how many moles are in 3.4*10^-7 grams of silicon dioxide, SiO2. I got 2.0^17 but I think it's wrong :(
blondinia [14]
Molar mass SiO2 = 28 + 32 = 60 

<span>so moles sand = 3.4 x 10-7 / 60</span>
8 0
3 years ago
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
2 years ago
Which statement about the electron-cloud model is true? It is the currently accepted atomic model. It can easily be replaced by
yan [13]

Answer:

3 choice

Explanation:

7 0
2 years ago
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