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ElenaW [278]
3 years ago
13

what quantity of ethylen glycol must be added to 125 g of water to raise the boiling point of 1.0 degrees celsuis

Chemistry
1 answer:
labwork [276]3 years ago
4 0

Answer:

We should add 15.15 grams of ethylen glycol.

Explanation:

Step 1: Data given

Mass of water = 125 grams

Temperature change = 1.0 °C

The boiling point elevation constant, Kbp, for water is 0.5121 °C/m

Step 2: Calculate mass needed

ΔT = i* Kb *  m

⇒ with i = the van't Hoff factor = 1

⇒ with Kb = 0.5121 °c/ kg/mol

⇒ with m = molality = moles / mass

ΔT = i* Kb *  m  ⇒ 1°C  = 0.5121 °C / kg/mol *   (X/0.125kg)

X = 0.2441 moles

Step 3: Calculate mass of ethylen glycol

Mass ethylen glycol = moles * molar mass

Mass ethylen glycol = 0.2441 moles * 62.068 g/mol

Mass of ethylen glycol = 15.15 grams

We should add 15.15 grams of ethylen glycol.

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Elements in the same group have the same number of valence electrons.

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For each element, predict where the "jump " occurs for successive ionization energies. (For example, does the jump occur between
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Answer:

A jump occurs when a core electron is removed.

Explanation:

A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.

For Beryllium, the electronic configuration of is 1s2 2s2.

There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron

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The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron

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8 0
3 years ago
A 26.08 g mixture of zinc and sodium is reacted with a stoichiometric amount of sulfuric acid. The reaction mixture is then reac
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Answer:

Molar percent of sodium in original mixture is 88,50%

Explanation:

The last reaction is:

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The moles of BaCl₂ are:

0,132L × 3,80M = 0,502 moles of BaCl₂

As the amount of BaCl₂ is the maximum possible to produce BaSO₄, the moles of BaCl₂ must be the same than moles of Na₂SO₄.

0,502 moles of BaCl₂ ≡ 0,502 moles of Na₂SO₄

These moles of Na₂SO₄ comes from:

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As the reaction is in stoichiometric amounts, moles of Na are twice the moles of Na₂SO₄

0,502 moles of Na₂SO₄ ×\frac{2molesNa}{1moleNa_{2}SO_{4}}× 22,99 g/mole = 23,08 g of Na

Molar percent of sodium in original mixture is:

\frac{23,08g}{26,08g}*100 = <em>88,50% </em>

I hope it helps

4 0
3 years ago
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