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2 years ago

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You are given a sample of limestone, which is mostly CaCO3, to determine the mass percentage of Ca in the rock. You dissolve the

limestone in hydrochloric acid, which gives a solution of calcium chloride. Then you precipitate the calcium ion in solution by adding sodium oxalate, Na2C2O4. The precipitate is calcium oxalate, CaC2O4. You find that a sample of limestone weighing 128.3 mg gives 140.2 mg of CaC2O4. What is the mass percentage of calcium in the limestone

2 answers:

irakobra [83]2 years ago

7 0

Answer:

34.15% is the mass percentage of calcium in the limestone.

Explanation:

Mass of precipitate that is calcium oxalate = 140.2 mg = 0.1402 g

1 mg = 0.001 g

Moles of calcium oxalate = $\frac{0.1402 g}{128 g/mol}=0.001095 mol$

1 mole of calcium oxalate have 1 mole of calcium atom.

Then 0.001095 moles of calcium oxalate will have 0.001095 moles of calcium atom.

Mass of 0.001095 moles of calcium :

0.001095 mol × 40 g/mol = 0.04381 g

Mass of sample of limestone = 128.3 mg = 0.1283 g

Percentage of calcium in limestone:

$\frac{0.04381 g}{0.1283 g}\times 100=34.15\%$

34.15% is the mass percentage of calcium in the limestone.

muminat2 years ago

6 0

Answer:

here are the percentages I got, I know this question is old but for others, for future reference, I figured I would add my answer. Hope this helps :)

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Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules.

kherson [118]

Answer:

The partial pressure of the other gases is 0.009 atm

Explanation:

Step 1: Data given

Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.

The atmospheric pressure = 0.90 atm

Step 2: Calculate mol fraction

If wehave 100 moles of air, 78 moles will be nitrogen,

21 moles will be oxygen, and 1 mol will be other gases.

Mol fraction = 1/100 = 0.01

Step 3: Calculate the partial pressure of the other gases

Pgas = Xgas * Ptotal

⇒ Pgas = the partial pressure = ?

⇒ Xgas = the mol fraction of the gas = 0.01

⇒Ptotal = the total pressure of the pressure = 0.90 atm

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3 years ago

Give the answer to the problem below using the correct number of significant digits. (1.3 x 103) x (5.724 x 104)

Gemiola [76]

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3 years ago

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The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl

nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1 x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

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Answer:

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2 years ago

Calculate the pH of a solution that is 0.235M benzoic acid and 0.130M sodium benzoate, a salt whose anion is the conjugate base

lesantik [10]

Hello!

We have the following data:

ps: we apply Ka in benzoic acid to the solution.

[acid] = 0.235 M (mol/L)

[salt] = 0.130 M (mol/L)

pKa (acetic acid buffer) =?

pH of a buffer =?

Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = $6.20*10^{-5}$

So:

$pKa = - log\:(Ka)$

$pKa = - log\:(6.20*10^{-5})$

$pKa = 5 - log\:6.20$

$pKa = 5 - 0.79$

$\boxed{pKa = 4.21}$

Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:

$pH = pKa + log\:\dfrac{[salt]}{[acid]}$

$pH = 4.21 + log\:\dfrac{0.130}{0.235}$

$pH = 4.21 + log\:0.55$

$pH = 4.21 + (-0.26)$

$pH = 4.21 - 0.26$

$\boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark$

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR! =)

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