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melisa1 [442]
3 years ago
5

Carbon disulfide is formed by the reaction of coke (carbon) with sulfur dioxide. how many moles of cs2 will be generated if 8.0

moles of coke react with a surplus of sulfur dioxide
Chemistry
1 answer:
trapecia [35]3 years ago
7 0
<span>The equation that produces carbon disulfide from the reaction of coke and sufur dioxide is expressed in the balanced equation: c+ 2sO2 = CS2 + 2O2. For every mole of coke reacted, there is one mole of carbon disulfide produced. Hence the answer here is 8 moles of CS2 </span>
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3 years ago
imagine that you go into the lab and perform a titration. you measure 40 ml of your analyte and add it to an erlenmeyer flask. t
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The Molar concentration of your analyte solution is 1.17 m

<h3>What is titration reaction?</h3>
  • Titration is a chemical analysis procedure that determines the amount of a sample's ingredient by adding a precisely known amount of another substance to the measured sample, with which the desired constituent reacts in a specific, known proportion.

Make use of the titration formula.

The formula is molarity (M) of the acid x volume (V) of the acid = molarity (M) of the base x volume (V) of the base.

if the titrant and analyte have a 1:1 mole ratio. (Molarity is a measure of a solution's concentration represented as the number of moles of solute per litre of solution.)

26 x 1.8 = 40 x M

M = 26 x1.8 /40

M = 1.17

The Molar concentration of your analyte solution is 1.17 m

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5 0
1 year ago
How does repeatability have an effect on scientific research?
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3 years ago
25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti
lesantik [10]

Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3.   This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

                HF  ⇄    H⁺    +      F⁻

C(eq)       [HF]     10⁻³M      0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M

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On the reaction side there are two or more reactants, and on the product side there is only one product. The reaction type is __
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