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tiny-mole [99]
2 years ago
12

NH4Cl + AgNO3 → AgCl + NH4NO3

Chemistry
1 answer:
iVinArrow [24]2 years ago
7 0
<h3><u>A</u><u> </u><u>N</u><u> </u><u>S</u><u> </u><u>W</u><u> </u><u>E</u><u> </u><u>R</u><u> </u><u>:</u><u> </u><u>–</u><u> </u></h3>

  • It's a balanced equation!
  • This is a precipitation reaction.
  • AgCl is the formed precipitate.
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based on your observation, you can infer that thermal energy transfers to/from objects with higher temperatures to/from objects
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Answer:

1. from      2. to

Explanation:

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3 years ago
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BlackZzzverrR [31]
2. answer is C.<span> Elements have the same physical and chemical properties in any period </span>
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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
Is heated water more or less dense than the melted snow And rainwater
san4es73 [151]

Answer:

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Explanation:

7 0
2 years ago
A compound is found to contain 33.3%
kakasveta [241]

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3 years ago
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