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3 years ago

A sample of sugar (C12H22O11) contains

Chemistry

1 answer:

Lady_Fox [76]3 years ago

5 0

4 moles of sugar.

Explanation:

A mole is defined as the amount of a substance contained in Avogadro's number of particles 6.02 x 10²³.

1 mole of substance = 6.02 x 10²³. molecules

Given that;

the sample of sugar contains 1.505 x 10²³.molecules

The number of moles in this amount of sugar is 4 moles

Learn more:

Number of moles brainly.com/question/13064292

#learnwithBrainly

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Calculate the volume (liters) of solution that can be prepared from 180. grams of K2Cr2O7 for a 0.200M solution.

Sladkaya [172]

Answer:

The volume of the solution is 0.305 liters.

Explanation:

Molar mass is the amount of mass that a substance contains in one mole. The molar mass of K₂Cr₂O₇ is 294 g / mole. Then you can apply the following rule of three: if by definition of molar mass 294 grams of the compound are contained in 1 mole, 180 grams are contained in how many moles?

$moles=\frac{180 grams*1mole}{294 grams}$

moles= 0.61

Molarity is a measure of the concentration of a substance that is defined as the number of moles contained in a certain volume. So, the molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

$molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

In this case:

molarity= 2 M
number of moles of solute= 0.61 moles
volume= ?

Replacing in the definition of molarity:

$2 M=\frac{0.61 moles}{volume}$

Solving:

$volume=\frac{0.61 moles}{2 M}$

volume= 0.305 liters

<u><em>The volume of the solution is 0.305 liters.</em></u>

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3 years ago

Which is longer, the carbon-oxygen single bond in a carboxylic acid or the carbon-oxygen bond in an alcohol?

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The carbon-oxygen single bond in a carboxylic acid

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150 ml of 0.1 m naoh is added to 200 ml of 0.1 m formic acid, and water is added to give a final volume of 1 l. what is the ph o

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Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles

Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles

Here, the limiting reagent is NaOH

The reaction is represented as:

HCOOH + NaOH ↔HCOONa + H2O

Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles

Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles

Total final volume is given as 1 L

Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M

[HCOONa] = 0.015/1 = 0.015 M

pKa of HCOOH = 3.74

As per Henderson-Hasselbalch equation

pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22

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