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vlada-n [284]
3 years ago
10

What lines up with south poles pointing toward the magnet

Chemistry
2 answers:
Delicious77 [7]3 years ago
8 0
Pretty sure its the north pole of the magnet
Olin [163]3 years ago
4 0
I think the answer is magnetic field?
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Predict the aldol product when the following ketone undergoes self-condensation in the presence of NaOH. Do not show the dehydra
Anit [1.1K]

Answer:

β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal.

Explanation:

When acetaldehyde is treated with dil.NaOH it undergoes self condensation as it contains alpha-hydrogen atom in its compound forming β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal. This compound upon further heating will eliminate a molecule of water forming aldol condensation product namely Crotonaldehyde Or But-2-en-al. see the diagram attached.

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3 years ago
What does an atom become when it gains an electron in an ionic bond?
Elden [556K]
It becomes a acceptor because in an ionic bonding the element who gives out is a donor while the atom which accept is a acceptor

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2 years ago
Do you think a battery system would OR wouldn’t have energy? If so, give me evidence/observations as to why you think so.
Basile [38]

Answer:

Yes it would

Explanation: Well it kinda depend on the voltage and how the battery has been in use or based on the condition

6 0
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Why is hydrogen a light gas and iron is a heavy solid
lbvjy [14]
<span>Heavier atoms make denser materials</span>
8 0
3 years ago
Read 2 more answers
Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0
2 years ago
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