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stich3 [128]
3 years ago
6

Calculate the molality of acetone in an aqueous solution with a mole fraction for acetone of 0.241. Answer in units of m.

Chemistry
1 answer:
Anit [1.1K]3 years ago
3 0

Answer: The molality of solution is 17.6 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molarity=\frac{n}{W_s}

where,

n = moles of solute

W_s = weight of solvent in kg

moles of acetone (solute) = 0.241

moles of water (solvent )= (1-0.241) = 0.759

mass of water (solvent )= moles\times {\text {Molar Mass}}=0.759\times 18=13.7g=0.0137kg

Now put all the given values in the formula of molality, we get

Molality=\frac{0.241}{0.0137kg}=17.6mole/kg

Therefore, the molality of solution is 17.6 mole/kg

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Answer:

The boiling point is the temperature at which the vapor pressure equals the pressure of gas.

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Explanation:

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3 years ago
Use the periodic table to determine the electron configuration for Ca and Pm in noble-gas notation Ca: [Ar]4s2 [Ar]4s1 [Ar]3s2 [
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Answer:

1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵

Explanation:

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Its atomic number is 20. So it has 20 protons and 20 electrons.

Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.

So, you have two more electrons (20 - 18 = 2) to distribute.

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Finally, the electron configuration is [Ar] 4s².

2) Pm

The atomic number of Pm is 61, so it has 61 protons and 61 electrons.

Pm is in the row (period) 6. So, the noble gas before Pm is Xe.

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The resultant distribution for Pm is: [Xe]6s² 4f⁵.
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What mass of C6H8O7 should be used every 7.0 X 10^2mg NaHCO3
Savatey [412]

Mass C₆H₈O₇ : 0.531484 g

<h3>Further explanation</h3>

Reaction

3NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + 3 H₂O (l) + Na₃C₆H₅O₇ (aq)

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mass NaHCO₃ : 7.10² mg=0.7 g

mol NaHCO₃ :

\tt mol=\dfrac{0.7}{84}=0.0083

mol C₆H₈O₇ :

\tt \dfrac{1}{3}\times 0.0083=0.00277

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mass C₆H₈O₇ :

\tt mass=0.00277\times 192=0.53184

4 0
3 years ago
If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN
viktelen [127]

Answer : The concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

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When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is SCN^- and Fe^{3+} is excess reagent.

First we have to calculate the moles of KSCN.

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\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M

Thus, the concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

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