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3 years ago

7

What is the maximum amount of Copper that could be converted to Cu2 ions by the concentrated nitric acid you used in part A of t

he experiment

1 answer:

Andreyy893 years ago

8 0

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Calculate the standard cell potential at 25 ∘c for the reaction x(s)+2y+(aq)→x2+(aq)+2y(s) where δh∘ = -793 kj and δs∘ = -319 j/

QveST [7]

First we will calculate free energy change:
ΔG₀ = ΔH₀ - (T * ΔS₀)
= - 793 kJ - (298 * - 0.319 kJ/K) = - 698 kJ
We know the relation between free energy change and cell potential is:
ΔG₀ = - n F E⁰ where
F = Faraday's constant = 96485 C/mol
n = 2 (given by equation that the electrons involved is 2)
ΔG₀ = - 2 x 96485 x E⁰
- 698 kJ = - 2 x 96485 x E⁰
E⁰ = (698 x 1000) / (2 x 96485) = 3.62 volts

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3 years ago

If 3.66•10^23 molecules of a substance have a mass of 24.31g, what is the molar mass of the substance?

BlackZzzverrR [31]

Answer is a I took the quiz

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2 years ago

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What is water clarity and why is it important?

il63 [147K]

Answer:

it's a measure of how far down light can penetrate through the water column. ... Because water clarity is closely related to light penetration, it has important implications for the diversity and productivity of aquatic life that a system can support

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3 years ago

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Which statement is TRUE about oxidation-reduction reactions? a. Every oxidation must be accompanied by a reduction. b. There are

Tomtit [17]

Answer:

Every oxidation must be accompanied by a reduction.

Explanation:

Oxidation and reduction are complementary processes. There can be no oxidation without reduction and vice versa. It is actually a given an take affair. A specie looses electrons which must be gained by another specie to complete the process. This explains why the selected option is the correct one.

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3 years ago

What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?

My name is Ann [436]

Answer:

602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The<em> yield</em> is measured by the amount of each product produced by the reaction.

The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

$2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O$

To calculate the yield follow these steps:

<u>1. Mole ratio</u>

$2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O$

<u />

<u>2. Convert 175mg of fluorene to number of moles</u>

$175mg/times 1g/1,000mg=0.175g$

Number of moles = mass in grams / molar mass

$\text{number of moles}=0.175g/166.223g/mol=0.0010528mol$

<u>3. Set a proportion for each product of the reaction</u>

a) <u>For CO₂</u>

i) number of moles

$2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x$

$x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2$

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

mass = number of moles × molar mass

mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) <u>For H₂O</u>

i) number of moles

$0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O$

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

mass = number of moles × molar mass

mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg

4 0

3 years ago