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BigorU [14]
2 years ago
10

The information below describes a redox reaction. 3 equations: First: upper C r superscript 3 plus (a q) plus 2 upper C l supers

cript minus (a q) right arrow upper C r (s) plus upper C l subscript 2 (g). Second: 2 upper C l superscript minus (a q) right arrow upper C l subscript 2 (g) plus 2 e superscript minus. Third: upper Cr superscript 3 plus (a q) plus 3 e superscript minus right arrow upper C r (s). What is the final, balanced equation for this reaction? 2 upper C r superscript 3 plus (a q) plus 6 upper C l superscript minus (a q) right arrow 2 upper C r (s) plus 3 upper C l subscript 2 (g). 2 upper C r superscript 3 plus (a q) plus 2 upper C l superscript minus (a q) plus 6 e superscript minus right arrow upper C l subscript 2 (g) plus 2 upper C r (s). Upper C r superscript 3 plus (a q) plus 6 upper C l superscript minus (a q) plus 3 e superscript minus right arrow 2 upper r (s) plus 3 upper C l subscript 2 (g). Upper C r superscript 3 plus (a q) plus 2 upper C l superscript minus (a q) right arrow upper C r (s) plus upper C l subscript 2 (g).
Chemistry
1 answer:
Marrrta [24]2 years ago
7 0

Answer:

A.

Explanation:

I just took the quiz.

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A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0
2 years ago
Which of the following pairs lists a substance that can neutralize HNO3 and the salt that would be produced from the reaction?
Nutka1998 [239]

NH₃, being a basic gas neutralizes the HNO₃ forming a salt NH₄NO₃

Therefore the correct answer is NH₃ and NH₄NO₃

The solution of which only 32% dissociates to release OH⁻ ions is a weak base. This is because some of the energy is used when the substance reacts with the solution thus some bonds are not broken.

HCl is an acid. This is because it dissociates  in water to give H⁺ as the only positively charged ions.

Arrhenius acid increases the concentration of hydrogen ions because it dissociates to release hydrogen ions as the only positively charged ions in the acid. So the answer is TRUE

Arrhenius base dissociates in water to release hydroxide ions as the only negatively charged ions.

NaOH⁺aq⇒Na⁺ ₍aq₎+ OH⁻₍aq₎

5 0
2 years ago
How many moles of N2 are essential for generating 0.08 moles of Li3N in the given reaction?6Li N2 2Li3N
jeka57 [31]
<span>08 moles Li3N * 1mole N2/2moles Li3N = 0.04 </span>
3 0
2 years ago
Read 2 more answers
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for
joja [24]

Answer:

ΔG° = -118x10³ J/mol

Explanation:

The two half-reactions in the cell are:

Oxidation half-reaction:

Co(s) → Co²⁺(aq) + 2e⁻; E° = -0,28V

Reduction half-reaction:

Cu²⁺(aq)+2e⁻ → Cu(s); E° = 0,34V

The E° of the cell is defined as:

E_{cell} = E_{red} - E_{ox}

Replacing:

0,34V - (-0,28V) = 0,62V

It is possible to obtain the keq from E°cell with Nernst equation thus:

nE°cell/0,0592 = log (keq)

Where:

E°cell is standard electrode potential (0.62 V)

n is number of electrons transferred (2 electrons, from the half-reactions)

Replacing:

0,62V×2/0,0592 = log (keq)

20,946 = log keq

keq = 8,83x10²⁰≈ 5,88x10²⁰

ΔG° is defined as:

ΔG° = -RT ln Keq

Where R is gas constant (8,314472 J/molK) and T is temperature (298K):

ΔG° = -8,314472 J/molK×298K ln5,88x10²⁰

<em>ΔG° = -118x10³ J/mol</em>

<em />

I hope it helps!

8 0
2 years ago
In an experiment, the density of an unknown liquid was calculated to be 0.78 g/ml. if the accepted value is 0.75 g/ml, calculate
Vikki [24]
Percentage error is the relative error your measured value is from the true or accepted value. The formula for percentage error is written below:

Percentage error = |True Value - Measured Value|/True Value   * 100
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Percentage error = 4%
6 0
1 year ago
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