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1 year ago

1)

1 answer:

5 0

Typically, the actual yield is lower than the theoretical yield because few reactions complete (i.e., are not 100% efficient) or because not all of the product in a reaction is collected. It is also conceivable for the real yield to exceed the theoretical yield.

What is theoretical yield ?

Theoretical yield is what would be obtained if 100% of the reaction was completed and no product was wasted in any way; however, procedures are not this efficient. Everything is dependent on the experimenter's precision and the reaction. Some reactions are reversible, meaning that less than 100% of the reaction proceeds to completion; some reactions do not go to completion because they require a large amount of energy or additional time, for example. Substances in the reaction may be lost throughout the method or difficult to separate from other sections of the experiment. In chemistry, some degree of inaccuracy is always expected.

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Calculate the mass in grams for a single molecule of carbon monoxide

Oxana [17]

Answer:

7.307 x 10^-23

Explanation:

mass of one molecule = mr/ avogadros number

= 44/6.022 x 10^23

= 7.307x 10^-23

4 0

3 years ago

Based on these data, what is the average atomic mass of element b? 10.01 10.51 10.81 11.01

Lelechka [254]

The right answer is C. 10.81

5 0

2 years ago

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The equilibrium constant for the dissolution of silver chloride (AgCl(s) Ag+(aq) + Cl–(aq)) has a value of 1.79 × 10–10. Which s

ryzh [129]

"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.

Option: b

<u>Explanation</u>:

As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant $\left(\boldsymbol{K}_{s p}\right)$ for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.

To calculate $\left(\boldsymbol{K}_{s p}\right)$ firstly molarity of ions are needed to be found with formula: $\text { Molarity of ions }=\frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}$

Then at equilibrium cations and anions concentration is considered same hence:

$\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }$

Hence from above data $\left(\boldsymbol{K}_{s p}\right)$ can be calculated by: $\left(\boldsymbol{K}_{s p}\right)$ = $\left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]$

6 0

3 years ago

Elements are organized on the periodic table based on their properties. Which statement correctly predicts and explains the chem

Elanso [62]

Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.

The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).

Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.

Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.

Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.

Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.

7 0

2 years ago

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Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)

Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation Cr(s) → Cr3+ + 3e-

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the half-reactions

2(Cr → Cr3+ + 3e-) E° ox = 0.74 V

3(Ni2+ +2e- → Ni) E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox = -0.25 + 0.74 = 0.49

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution) MnO4- +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e- (oxidation) Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the half-reactions

MnO4- +8H+ +10-e- ⇔ Mn2+ +4H2O E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e- E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox = 1.51 + 2.12 = 3.63 V

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0

2 years ago