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2 years ago

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Deshocribe how you could separate and purify compound A from a mixture of two neutral compounds (A and B) when A comprises 95% o

f the total and B the other 5% of the total. Assume that A and B have similar polarities.

1 answer:

uranmaximum [27]2 years ago

3 0

Answer:

Good Luck!

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astrioids can be found at the astrioid belt

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2 years ago

A sample was prepared by mixing 18. ml of 3.00 x 10^-3 m crystal violet (cv) with 2.00 ml of 0.250 m naoh. calculate the resulti

Aleks [24]

Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.

Explanation :

Step 1 : Find moles of crystal violet and NaOH.

The molarity formula is

$Molarity = \frac{mol}{L}$

Molarity of crystal violet = $3.00 \times 10^{-3} = \frac{mol (CrystalViolet)}{L}$

The volume of crystal violet solution is 18 mL which is 0.018 L.

Moles of crystal violet = $3.00 \times 10^{-3} \times 0.018 = 5.4 \times 10^{-5}$

Moles of crystal violet = 5.4 x 10⁻⁵

Moles of NaOH = $Molarity \times L = 0.250 \times 0.00200 = 5.00 \times 10^{-4}$

Moles of NaOH = 5.00 x 10⁻⁴

Step 2 : Find total volume of the solution

The total volume of the solution after mixing NaOH and crystal violet is

0.018 L + 0.00200 = 0.020 L

Step 3 : Use molarity formula to find final concentrations

Molarity of crystal violet = $\frac{mol(CrystalViolet)}{Total Volume(L) } = \frac{5.4 \times 10^{-5}}{0.020} = 2.7 \times 10^{-3}$

Final concentration of CV = 0.0027 M

Molarity of NaOH= $\frac{mol(NaOH)}{Total Volume(L) } = \frac{5.00 \times 10^{-4}}{0.020} = 0.025 \times 10^{-3}$

NaOH is a strong base and dissociates completely as follows.

$NaOH (aq) \rightarrow Na^{+} (aq) + OH^{-} (aq)$

The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.

Concentration of OH⁻ = 0.025 M

8 0

3 years ago

What is the molar ratio for the following equation after it has been properly balanced?

hichkok12 [17]

6 Na + 1 Fe₂O₃ → 3 Na₂O + 6 Fe

<h3>Explanation</h3>

Method One: Refer to electron transfers.

Oxidation states:

Na: from 0 to +1; loses one electron.
Fe: from +3 to 0; gains three electrons.

Each mole of Fe₂O₃ contains two Fe atoms and will gain 2 × 3 = 6 electrons during the reaction. It takes 6 moles of Na to supply all those electrons.

6 Na + 1 Fe₂O₃ → ? Na₂O + ? Fe

There are two moles of Na atoms in each mole of Na₂O. 6 moles of Na will make 3 moles of Na₂O.
There are two moles of Fe atoms in each mole of Fe₂O₃. 1 mole of Fe₂O₃ will make 2 moles of Fe.

6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe

Method Two: Atoms conserve.

Fe₂O₃ has the largest number of atoms among one mole of all four species in this reaction. Assume <em>one</em> as its coefficient.

? Na + <em>1</em> Fe₂O₃ → ? Na₂O + ? Fe

There are two moles of Fe atoms and three moles of O atoms in each mol of Fe₂O₃. One mole of Fe₂O₃ contains two moles of Fe and three moles of O. There are one mole of O atom in every mole of Na₂O. Three moles of O will go to three moles of Na₂O.

? Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe

Each mole of Na₂O contains two moles of Na. Three moles of Na₂O will contain six moles of Na.

<em>6</em> Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe

Simplify the coefficients. All coefficients in this equation are now full number and relatively prime. Hence the equation is balanced.

6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe

6 0

3 years ago