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2 years ago

analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula

2 answers:

6 0

1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = 0.0135 mol Cr
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr /0.0135= 1 mol Cr
0.0538 mol O/0.035= 4 mol Cr
K2CrO4

Sauron [17]2 years ago

3 0

Answer: The empirical formula for the given compound is $K_2CrO_4$

Explanation:

We are given:

Mass of Cr = 0.70 g

Mass of K = 1.04 g

Mass of O = 0.86 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Chromium = $\frac{\text{Given mass of Chromium}}{\text{Molar mass of Chromium}}=\frac{0.70g}{52g/mole}=0.013moles$

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{1.04g}{39g/mole}=0.027moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.86g}{16g/mole}=0.054moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.013 moles.

For Chromium = $\frac{0.013}{0.013}=1$

For Potassium = $\frac{0.027}{0.013}=2.07\approx 2$

For Oxygen = $\frac{0.054}{0.013}=4.15\approx 4$

Step 3: Taking the mole ratio as their subscripts.

The ratio of K : Cr : O = 2 : 1 : 4

Hence, the empirical formula for the given compound is $K_2CrO_4$

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Explanation:

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The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add

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Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

Step-by-step solution:

To solve this:

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

The solubility of KCl given is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

The Ksp was found to be equal to 14.

In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x= molar solubility [K+],/[Cl-] :. × , x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b= molar solubility :. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

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