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Marta_Voda [28]
2 years ago
11

freezing a sealed full bottl of water leads to a broken bottle and freezing a sealed full flexible wall container of n butyl alc

holol leads to a container with concave walls, what can you conclude about the densitiies of the solid and liquid forms of these two substances
Chemistry
1 answer:
Nataly [62]2 years ago
4 0

Answer:

See explanation

Explanation:

We know that water begins to expand at 4°C. This is generally referred to as the anomalous behavior of water.

The implication of this is that ice has a greater volume than water. Thus ice is less dense than liquid water.

n-butyl alcohol has much less density than water. This accounts for the fact that water upon freezing breaks the bottle while n butyl alcholol leads to a container with concave walls.

Hence the density of ice is much higher than the density of n butyl alcohol.

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A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing point depression of
sergij07 [2.7K]
From the equation; ΔTf = Kf × m
Where, Kf for water = 1.853 K kg/mole; m is the molarity = number of solute/amount of solvent in kg.
Glucose is the solute whose molecular mass is 180 g/mole and water is the solvent. 
Moles of solute = 15.5/180 = 0.0861 moles
Amount of solvent in kg = 245/1000 = 0.245 Kg
Therefore; molarity = 0.0861/0.245 = 0.3515 moles/Kg
Therefore; ΔTf = 1.853 × 0.3515 = 0.6513 K
Hence; the depression in freezing point is 0.6513 
The freezing point of solution will therefore be;
= 273 - 0.6513 = 272.3487 K

7 0
3 years ago
Which of these common substances is a homogeneous mixture?
IgorC [24]

Answer:

whole milk ,table salt , maple syrup

3 0
2 years ago
Read 2 more answers
Which element in each of the following sets would you expect to have the lowest IE₃?
AysviL [449]

Answer:

(a) AL

(b) Sc

(c)Al

Explanation:

Ionization Energy is the energy required to remove electrons from the outer most orbitals of atom.

The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.

By looking at electron configuration we can figure out which electron will need more energy.

<h3>(a)Na, Mg, Al</h3>

1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹

Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

<h3>(b) K, Ca, Sc</h3>

K₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

<h3>(c) Li, Al, B</h3>

Li₃ ⇒ 1s², 2s¹

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

B₅ ⇒ 1s², 2s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

3 0
3 years ago
Read 2 more answers
Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc
Anni [7]

Answer:

Ka = 4.76108

Explanation:

  • CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

⇒ Ka = 4.76108

7 0
3 years ago
Can anyone help me with this?
Scrat [10]

Explanation:

c2H5

|

cH3 _ c _ c H2 _ c H3

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c2H5

cH3

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cH3_ cH _ cH_ cH2_ cH3

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cH3

5 0
2 years ago
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