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34kurt
3 years ago
15

Balance the following chemical equation:NaBr+Cl2, –NaCl+Br2​

Chemistry
1 answer:
svlad2 [7]3 years ago
6 0

Hey there!

NaBr + Cl₂ → NaCl + Br₂

First balance Na.

One on the left, one on the right: already balanced.

Next balance Br.

One on the left, two on the right. Add a coefficient of 2 in front of NaBr.

2NaBr + Cl₂ → NaCl + Br₂

Lastly balance Cl.

Two on the left, one on the right. Add a coefficient of 2 in front of NaCl.

2NaBr + Cl₂ → 2NaCl + Br₂  

This is our final balanced equation.

Hope this helps!

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Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

<h3>What is the boiling-point elevation?</h3>

Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

  • Step 1: Calculate the molality of the solution.

We will use the definition of molality.

b = mass solute / molar mass solute × kg solvent

b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m

  • Step 2: Calculate the boiling-point elevation.

We will use the following expression.

ΔT = Kb × m × i

ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C

where

  • ΔT is the boiling-point elevation
  • Kb is the ebullioscopic constant.
  • b is the molality.
  • i is the Van't Hoff factor (i = 2 for NaCl).

The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:

100 °C + 0.140 °C = 100.14 °C

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

Learn more about boiling-point elevation here: brainly.com/question/4206205

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